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 1 answerLast reply by: Professor Selhorst-JonesMon Aug 3, 2015 5:54 PMPost by Mike Petersen on July 3, 2015Would this be considered the "squeeze theorem"? 1 answerLast reply by: Professor Selhorst-JonesMon Jul 21, 2014 8:29 PMPost by Teemu S on July 20, 2014What an awesome lecture! I know I'm going have to deal with this definition in a couple of months and this has cleared it up for me. I really appreciate all the lectures you've given. You are by far the most motivating and inspiring math teacher I've come across. A big thank you from Finland! 2 answersLast reply by: Jamal TischlerSun Mar 1, 2015 7:09 AMPost by Jason Todd on July 27, 2013Awesome Video!!! It grabbed my intuition and slammed it into synchrony with this concept. Thank you! The dialogue made things really enjoyable. P.S. f(x)= x sin(1/x) is my new favorite function! Do you have a list of really cool functions like this?

### Formal Definition of a Limit

Note: Very few students will have any use for what is in this lesson, let alone during a pre-calculus level course. Problems based on the formal definition of a limit are extremely rare in calculus class. This stuff won't come up in science classes, and will only be necessary in high-level college math. Still, if you're interested in math for math's sake or know that you want to one day study higher level mathematics, check this lesson out. It's okay if it doesn't all make perfect sense today, but seeing these ideas now will help you down the road when it starts mattering. Plus, it's fascinating stuff!
• We need two new greek letters:    Delta⇒ δ    Epsilon⇒ ε
• Formal Definition of a Limit: Let f be a function defined on some interval (a,b) of the real numbers, where a < b. Let c ∈ (a,b) and L be a real number. Then,
 lim f(x)   =  L
means that for any real ε > 0, there exists some real δ > 0 such that for all x where 0 < |x−c| < δ, we have |f(x) − L| < ε.
• The first half of the definition sets up the groundwork. The function is defined, there is a horizontal location we're approaching (c), and some vertical location the function goes towards (L).
• The second half of the definition is the difficult part. It says that for any ε > 0 boundary around L, we can restrict our x to within some δ > 0 boundary of c to force f(x) to stay within the ε-boundary around L.
• Check out the video to see some diagrams and more explanation of this idea. It's a confusing idea at first, and it takes awhile to make sense of it.
• We can imagine a limit as a sort of never-ending debate between two people. One person gives hypothetical ε-boundaries, while the other person has to defend by giving a δ-boundary that would keep the function within the ε-boundary.
• To formally prove a limit exists, we must show that for any ε > 0, there will always be some δ > 0. This normally takes the form of creating a formula for δ based off ε, because if we can create such a formula, then we will have shown δ will exist no matter what ε is chosen.

### Formal Definition of a Limit

Prove the below limit using the formal definition of a limit.
 limx→ −2 1−x    =  3
• (Note: As is discussed in the video lesson, the formal definition of a limit is extremely precise and technical. Not many students will have a serious need for the formal definition of a limit: it is extremely unlikely to show up during a high school course, and will only start showing up in high-level college math courses. Unless you plan to specifically pursue higher-level mathematics, understanding this concept perfectly will not affect your grades or performance in other classes and fields of study. That said, this stuff can be fascinating. If you find this kind of formal proof and high-level puzzle solving interesting and engaging, you might want to consider studying higher-level mathematics. The specific field of math that these sort of puzzle-ideas show up in is called abstract or pure mathematics. You can find courses full of this kind of material in college. Even if you can't attend such courses at college, you can study the material on your own if you're interested. A good start would be an introductory text to a branch of mathematics like Number Theory or Combinatorics: these fields can be quite accessible to beginners, but still are based on similar ideas of high-level conceptual thought and analytic proof/puzzle-solving.)
• Understand the Approach: When working on a proof, it's necessary to first understand how you will prove the claim. Like writing an essay, you want some sort of outline-a skeleton to build on-before you start writing the proof. Come up with a brief proof sketch and some notes that make sense to yourself for whatever you're trying to prove. For this problem we need to show that, for any ε > 0, there will always exist some δ > 0 that will fulfill the requirements of the formal epsilon-delta limit definition. That means we must begin by figuring out a method to create δ in general from an arbitrary ε value. Once we have such a method, we then prove the limit's existence by showing that the method works.
• Setting Up to Find δ: To help us see what we're working on, compare the limit this problem is about to the formal structure in the limit defintion:
 limx→ c f(x)   =  L        ⇔ limx→ −2 1−x    =  3
Thus, for this problem, we have c=−2,  f(x) = 1−x, and L = 3. This means, for this problem, what we will be trying to show is that for ε > 0, there exists some δ > 0 such that
 for all x where   0 < |x−c| < δ,    we have that   |f(x) − L|<ε.
Or, plugging in the specific values we're using, we have
 for all x where   0 < |x−(−2)| < δ,    we have that   |f(x) − 3|<ε.

Therefore we need a way to figure out δ given any value for ε. To figure this out, we can pretend as if we knew the δ, set it up, then solve the resulting relation for what δ must have been in the first place.
• Calculating a Method for δ: Notice that, since we must have 0 < |x−c| < δ, this implies that
 x is contained in the interval of (c− δ,  c+δ).
Because of the function we're working on (f(x) = 1−x), notice that the most extreme values we can obtain for f(x) would be right at the edge of using either x=c−δ or x=c+δ [technically, they're not contained in the interval, but they give us the "boundary"]. With this in mind, let us consider using x=c−δ in the function. Notice two things: 1) Since x=c−δ is the "boundary" of what x can be, the value of f(x) will be right at the edge value for what it is allowed to be ⇒ L±ε.   2) Since f(x) = 1−x, a value lower than c (like c−δ) will actually produce an increase in the value of f(x): this means we can figure out δ using
 f(c−δ) = L +ε    ⇒     1 − ⎛⎝ −2 − δ ⎞⎠ = 3+ ε
With this equation, we can now solve for a general way to find δ from knowing ε:
 1 − ⎛⎝ −2 − δ ⎞⎠ = 3+ ε    ⇒     3 + δ = 3 + ε    ⇒     δ = ε
Thus, for this problem, it is enough to set δ = ε.

[Remarks: This part is almost certainly the hardest part about setting up a formal limit proof. It can be confusing to come up with a method to create δ. If you find it difficult to follow the above or particularly hard to do on your own, just try playing around with plugging in x=c±δ and get a sense for how the function works around x=c. Once you have a sense for that, you can probably guess at a reasonable choice for setting up δ from ε. Test out your guess, and if it works, use it in the proof. If it doesn't work for some reason, you'll get a better sense for how the function works and you can use that to come up with a new way to set up δ. Also, notice that the above is not part of the proof itself. It is a step we do outside of the proof as preparation. It's like a magician doing a magic-hat trick: before we start, we put the rabbit in the hat. That way, once we start the performance of the proof, we can pull the rabbit out of the hat at the right moment.]
• Proof: We claim that limx→ −2  1−x    =  3. We show this by proving that for any ε > 0, there exists some δ > 0 that will satisfy the formal definition of a limit-that is, |f(x) − 3|<ε for any x where 0 < |x−(−2)| < δ.

Consider any possible value of ε such that ε > 0. Let δ = ε. We now show that this δ will always satisfy the definition of a limit. Notice that, because we are working with the function f(x) = 1−x, for any number a where a > 0, the below is true:
 f(x+a)   <  f(x)   <  f(x−a)
[This is because  f(x+a) = 1 −x −a    and    f(x−a) = 1 −x +a.] Furthermore, the larger the value of a, the greater the distance between f(x±a) and f(x) becomes. With this in mind, notice that since 0 < |x−(−2)|<δ, we have that x is bounded as
 −2 − δ  <  x   <  −2 + δ.
Also, since ε > 0 and we set δ = ε, we have that δ > 0. Thus, by the same logic that gave us our observation about f(x±a), we have that for any x that is bounded by the above inequality, the below is true
 f(−2 + δ)   < f(x)   < f(−2 − δ).
Evaluate the left and right functions based on their inputs:
 1 − (−2 + δ)
 <
 f(x)
 <
 1 − (−2 − δ)
 1 +2 − δ
 <
 f(x)
 <
 1 +2 + δ
 3 − δ
 <
 f(x)
 <
 3 + δ
 − δ
 <
 f(x)−3
 <
 δ
By the nature of absolute value, the above is equivalent to
 |f(x) − 3|<δ.
Finally, since we originally set δ = ε, we have
 |f(x) − 3|<ε,
which is what we set out to show. Therefore the method for choosing δ will work for any ε, so we have proven that the limit exists.
Proof. [By the nature of proof, it cannot be given as a short answer. See the final step to the problem for a detailed proof, and the preceding steps to help understand some of the motivations for how the proof is set up.]
Let f(x) = {
 2,
 x < 1
 3,
 x ≥ 1
.    Formally prove that limx→ 1  f(x) does not exist.
• (Note: As is discussed in the video lesson, the formal definition of a limit is extremely precise and technical. Not many students will have a serious need for the formal definition of a limit: it is extremely unlikely to show up during a high school course, and will only start showing up in high-level college math courses. Unless you plan to specifically pursue higher-level mathematics, understanding this concept perfectly will not affect your grades or performance in other classes and fields of study. That said, this stuff can be fascinating. If you find this kind of formal proof and high-level puzzle solving interesting and engaging, you might want to consider studying higher-level mathematics. The specific field of math that these sort of puzzle-ideas show up in is called abstract or pure mathematics. You can find courses full of this kind of material in college. Even if you can't attend such courses at college, you can study the material on your own if you're interested. A good start would be an introductory text to a branch of mathematics like Number Theory or Combinatorics: these fields can be quite accessible to beginners, but still are based on similar ideas of high-level conceptual thought and analytic proof/puzzle-solving.)
• Understand the Approach: When working on a proof, it's necessary to first understand how you will prove the claim. Like writing an essay, you want some sort of outline-a skeleton to build on-before you start writing the proof. Come up with a brief proof sketch and some notes that make sense to yourself for whatever you're trying to prove. For this problem, we need to show that the limit can not exist. That is, we need to show there exists some ε > 0 such that there is no possible δ that will fulfill the requirements of the formal epsilon-delta limit definition. If we can find some ε and show that it is absolutely impossible to satisfy it, we will have completed the proof.
• Working out a "broken" value for ε: To figure out a value of ε that will be impossible to satisfy in the limit definition, we first need to understand how the function works. A great way to do this is by drawing a graph. Graph the function, and focus on the area around the location we're interested in (x→ 1). [Remember, f(x) is a piecewise function in this problem. If you're unfamiliar with them, check out the lesson Piecewise Functions.] Looking over the graph, we see that the reason the limit will not exist is because of the "jump" at x=1. We will prove the nonexistence of the limit by showing that the "jump" makes it impossible. If we choose the right ε, we can see that no δ can satisfy it because x=1 is situated directly on the "jump". At this point, remember what ε represents: |f(x) − L|<ε. In other words, all the points in the "neighborhood" of the limit location must be less than ε distance away from whatever L is. Notice that the best compromise we could make for an L would be if it were halfway between 2 and 3. But then it would still have a distance of [1/2] between it and each part of the graph. This realization allows us to find a value for ε that must fail:
 ε = 1 2
This value will force the limit to fail, because if it is less than a distance of [1/2] from one of the two chunks in the graph, it must be more than [1/2] from the other chunk. Thus it is impossible to fulfill the formal limit definition using this value for ε. [Remarks: While the above value will work to show the limit is impossible, it is not the only value we could use. As long as we chose any ε such that 0 < ε ≤ [1/2], the formal limit definition would be impossible. Also, notice that figuring out an appropriate value for ε is not part of the proof itself. It is a step we do outside of the proof as preparation. Once we know a value we can use, we then present it in the proof, but figuring it out is something we do beforehand.]
• Proof (by contradiction): Since we're trying to prove that something is impossible, a good way to show this is by initially assuming it is true, then making it clear that if it were true, it would be nonsense and contradict itself. We will proceed by using this method of contradiction.

Assume the limit does exist. That is, there is some L where limx→ 1  f(x)=L. [Note: We are not saying we know what the value of L is-all we're saying is that the limit does exist and that there must be some L out there that is the value of the limit.] Thus, by the formal limit definition, it must be that for any ε > 0, there exists some δ > 0 such that for any x where 0 < |x−1| <δ, we have |f(x) − L|<ε. With this requirement in mind, consider ε = [1/2]. By the limit definition, there must exist some δ > 0 that will satisfy |f(x) − L|< [1/2]. Now notice that, no matter what the value of δ is, it must be the case that there exist some numbers a and b such that
 1−δ < a < 1        and        1 < b < 1+δ
[This is because δ > 0, so there will always be room to "fit" in some a and some b on either side of 1.] Notice that a and b satisfy the same requirement that x is bounded by. Because we know that where 0 < |x−1| <δ, we have |f(x) − L|< [1/2], it must also be the case that
 |f(a) − L|< 1 2 and        |f(b) − L|< 1 2
Let's now evaluate the functions. Since a < 1 and b > 1, the piecewise function f(x) gives us the below when we evaluate:
 |2 − L|< 1 2 and        |3 − L|< 1 2
Let us now make the restrictions on L clear to ourselves. By the definition of absolute value, we have
 − 1 2 < 2 − L < 1 2 and        − 1 2 < 3 − L < 1 2
Which, through some simple algebra, we can arrange to get
 3 2 < L < 5 2 and 5 2 < L < 7 2
But this is clearly impossible! There is no value L that could possibly satisfy both of the above inequalities simultaneously.

The above realization gives us a contradiction! We arrived at this impossible conclusion through sound logic, so the only possible explanation is that our initial assumption (the existence of the limit) was wrong. Therefore the limx→ 1  f(x) does not exist.
Proof. [By the nature of proof, it cannot be given as a short answer. See the final step to the problem for a detailed proof, and the preceding steps to help understand some of the motivations for how the proof is set up.]
Prove the below limit using the formal definition of a limit.
 limx→ 4 − 20 x = −5
• (Note: As is discussed in the video lesson, the formal definition of a limit is extremely precise and technical. Not many students will have a serious need for the formal definition of a limit: it is extremely unlikely to show up during a high school course, and will only start showing up in high-level college math courses. Unless you plan to specifically pursue higher-level mathematics, understanding this concept perfectly will not affect your grades or performance in other classes and fields of study. That said, this stuff can be fascinating. If you find this kind of formal proof and high-level puzzle solving interesting and engaging, you might want to consider studying higher-level mathematics. The specific field of math that these sort of puzzle-ideas show up in is called abstract or pure mathematics. You can find courses full of this kind of material in college. Even if you can't attend such courses at college, you can study the material on your own if you're interested. A good start would be an introductory text to a branch of mathematics like Number Theory or Combinatorics: these fields can be quite accessible to beginners, but still are based on similar ideas of high-level conceptual thought and analytic proof/puzzle-solving.)
• Understand the Approach: When working on a proof, it's necessary to first understand how you will prove the claim. Like writing an essay, you want some sort of outline-a skeleton to build on-before you start writing the proof. Come up with a brief proof sketch and some notes that make sense to yourself for whatever you're trying to prove. For this problem we need to show that, for any ε > 0, there will always exist some δ > 0 that will fulfill the requirements of the formal epsilon-delta limit definition. That means we must begin by figuring out a method to create δ in general from an arbitrary ε value. Once we have such a method, we then prove the limit's existence by showing that the method works. On a challenging problem like this one, it will help us to think carefully about how the function works and to draw a graph of it. Doing so will allow us to consider how δ affects the range of outputs the function can have, and thus give us a better sense of how we should set our method to choose δ from ε.
• Setting Up to Find δ: To help us see what we're working on, compare the limit this problem is about to the formal structure in the limit defintion:
 limx→ c f(x)   =  L        ⇔ limx→ 4 − 20 x = −5
Thus, for this problem, we have c=4,  f(x) = −[20/x], and L = −5.

Begin by noticing that there is a potential complication for f(x): if we ever allow x=0, the function will break down because we will divide by 0. With this in mind, we need to make sure our δ never allows this to happen. Since c=4, this means the x-values must be bound within a distance of 4, so we must have a restriction of δ ≤ 4. For ease, let us cap the largest possible δ at 1. Whatever δ choosing method we come up with below, we will always choose δ so that we have δ ≤ 1. This will keep us safe from accidentally choosing a δ large enough to allow for dividing by 0 in f(x). [We could cap δ at 4, or any other smaller number, but 1 is nice and round, so we'll choose that.] [Remark: The above will actually turn out to be unnecessary. In the next step, we will see that our method of choosing δ from ε will wind up forcing δ < 4, so explicitly capping δ so it never goes above that value will not be needed. Still, this is an important idea in general and can be useful in other problems. Capping a largest possible size for the δ (as is done in the video lesson) can sometimes give you useful restrictions that make it easier to fulfill the requirements of the epsilon-delta limit definition.]
• Setting Up to Find δ, cont.: Now let's start working out a method to find δ based on the value of ε. Notice that, in general (but not always), we can solve for δ in both of the below inequalities to figure out how to restrict δ:
 L− ε ≤ f(c− δ) ≤ L + ε       and        L− ε ≤ f(c+ δ) ≤ L + ε
Of course, that means an awful lot of algebra, and results in many different inequalities that we have to sort through to figure out how to choose δ. It usually works, but we want something a little bit easier and faster.

To come up with an easier way to figure out how to choose δ, we want to carefully think about the specific function we're working with. Draw a graph of the function (which you can see in red below). Now notice how going left or right of c=4 by δ will affect the output of f(x): Looking at the graph, we see that f(4−δ) will be farther away from L=−5 than f(4+δ), no matter what δ we choose. Furthermore, we see that in the interval of [4−δ,  4+δ], the point of x=4−δ gives the biggest height difference from −5. Thus, with the above in mind, we see that if f(4−δ) is at the ε boundary "edge" from −5 (that is, −5−ε), all of the x-values that are less than δ from 4 will have to be within the ε boundary. This means we can figure out the largest possible δ by solving the below equation:
 f(4−δ) = −5−ε
Work out the largest possible δ:
 − 20 4−δ = −5 −ε    ⇒     − 20 −5−ε = 4−δ    ⇒ 20 5+ε = 4−δ    ⇒     δ = 4− 20 5+ε
We will use the boxed portion above as our method of choosing δ from ε. Working it out has also given us an added benefit: notice that we never have to worry about δ ≥ 4. Since ε > 0, there is no ε that can possibly cause us to choose δ = 4 or anything larger, so we're safe from accidentally "breaking" f(x). This means we can discard the requirement we came up with in the previous step of making sure δ is always smaller than some value. The reason for that previous requirement will never show up because of how we will be choosing δ from ε. Furthermore, we don't have to worry about our method of choosing δ being able to "break" either: because ε > 0, there is no way [20/(5+ε)] can have a denominator of 0 so we can be guaranteed that δ will be defined. Thus it is enough to simply set δ based on the boxed portion above.

[Remarks: This part is almost certainly the hardest part about setting up a formal limit proof. It can be confusing to come up with a method to create δ. If you find it difficult to follow the above or particularly hard to do on your own, just try playing around with plugging in x=c±δ and get a sense for how the function works around x=c. Once you have a sense for that, you can probably guess at a reasonable choice for setting up δ from ε. Test out your guess, and if it works, use it in the proof. If it doesn't work for some reason, you'll get a better sense for how the function works and you can use that to come up with a new way to set up δ. Also, notice that the above is not part of the proof itself. Once we know a value we can use, we then present it in the proof, but figuring it out is something we do beforehand.]
• Proof: We claim that limx→ 4  −[20/x] = −5. We show this by proving that for any ε > 0, there exists some δ > 0 that will satisfy the formal definition of a limit-that is, |f(x) − (−5)|<ε for any x where 0 < |x−4| < δ.

Consider any possible value of ε such that ε > 0. We set δ as the below
 δ = 4− 20 5+ε ,
and we will now show that setting δ in such a manner will always satisfy the definition of a limit.

Begin by noticing that we will always have δ > 0 (as is necessary for the definition of a limit)-this comes from the fact that ε > 0. Furthermore, because ε > 0, it must be that δ < 4. This means we do not have to worry about any potential issues that would be created by the vertical asymptote at x=0 because x=0 can never be in the interval implied by 0 < |x−4| < δ.

Next, notice that for any x where we have 0 < |x−4| < δ, that condition requires x to be contained in the interval of (4−δ,  4+δ). Because of how the function f(x) works (see the graph in the above step), it must be the case that for any value of x contained in the interval, we have
 f(4−δ)
 <
 f(x)
 <
 f(4+δ)
Evaluate the left and right functions based on their inputs, then plug in the value we determined for δ:
 − 20 4−δ <     f(x) <     − 20 4+δ

20

 4− ⎛⎝ 4− 20 5+ε ⎞⎠
<     f(x) <     − 20

 4+4− 20 5+ε
We now simplify to better understand the inequality:
20

 20 5+ε
<     f(x) <     − 20

 8− 20 5+ε

 − 20(5+ε) 20 <     f(x) <     − 20(5+ε) 8(5+ε)−20

 −(5+ε)     <     f(x) <     − 100+20ε40+8ε−20

 −5−ε    <     f(x) <     − 100+20ε20+8ε
At this point, the left side of the above inequality shows what we want (Remember, our goal is to show that |f(x) − (−5)|<ε, or, equivalently, −5−ε  <  f(x)   <  −5+ε.), but the right side of the inequality is rather confusing. We want to show that f(x)   <  −5 + ε, but it is hard to see if that is true. To show that this is the case, we claim the below to be true:
 − 100+20ε20+8ε <   −5 + ε
Now we'll prove that our claim is true through some basic algebra:
 − 100+20ε20+8ε <   −5 + ε

 −(100+20ε)    <   (−5 + ε)(20+8ε)

 −100 −20 ε   <   −100 −20 ε+ 8 ε2

 0    <   8 ε2
Since the final inequality is true (we know ε > 0, so it must be true) and each inequality is equivalent to the one above it through algebra, we have now shown that our claim is true:
 − 100+20ε20+8ε <   −5 + ε
Therefore, by combining this with the result we arrived at previously, we have
 −5−ε    <     f(x)     <     − 100+20ε20+8ε <     −5 + ε,
and so
 −5−ε    <     f(x)     <     −5 + ε,
which, by the nature of absolute value, means that
 |f(x) − (−5) |<ε,
which is what we set out to show. Therefore the method of choosing δ will work for any ε, so we have that the limit exists.
Proof. [By the nature of proof, it cannot be given as a short answer. See the final step to the problem for a detailed proof, and the preceding steps to help understand some of the motivations for how the proof is set up.]
When functions interact inside of a limit, they often interact as we would expect them to do so. For example, the addition law of limits says the below:
 "If limx→ a f(x) =L and limx→ a g(x) =M, then limx→ a ⎛⎝ f(x) + g(x) ⎞⎠ =L+M."

Thinking along these lines, we would expect there to be a composition law for limits that says the below:
 "If limx→ a f(x) =b and limx→ b g(x) =y, then limx→ a g ⎛⎝ f(x) ⎞⎠ =y."
However, the above claim (the composition "law") is not true. Show why this is false by giving a counter-example to the above claim that uses the two functions below:
f(x) = 1,              g(x) =

 x,
 x ≠ 1
 2,
 x = 1
• Start off by understanding how the (fake) composition "law" is supposed to work. We have
 limx→ a f(x) = b        and limx→ b g(x) = y,
and we're interested in looking at
 limx→ a g ⎛⎝ f(x) ⎞⎠ .
Well, we would probably expect that since limx→ a f(x) = b, we would have
 limx→ a g ⎛⎝ f(x) ⎞⎠ =     g ⎛⎝ limx→ a f(x) ⎞⎠ = limf* → b g(f*)     =     y
The above is the line of thinking that the composition "law" is based on.

However!, the problem told us that it does not work, and that we are going to show a counter-example of when it fails. Let's look into the functions that the problem told us to use for the counter-example.
• First, look at f(x) = 1. This is just a constant function: no matter what x-value we use as input, the output is always 1. Next, take a look at g(x). This one is a piecewise function, so it's a little more complicated. Let's take a look at its graph to better understand it. We see that there is a small "hole" in the line where the function "jumps" up to f(1) = 2. However, the jump only occurs at this single location.
• While the problem told us the f(x) and g(x) we should use for creating the counter-example, it told us nothing about the locations used in the limits the "law" is based on:
 limx→ a f(x) = b        and limx→ b g(x) = y,
Specifically, the locations x→ a and x→ b. Now we need to figure out what values we should use for a and b. Deal with f(x) first. Since f(x) = 1 is a constant function, it will have the same limit everywhere: 1. The value we choose for a is totally unimportant, so just pick some random number. Let's use a=47.
 limx→ a f(x) = b     ⇒ limx → 47 f(x) = 1
This also makes it quite easy to figure out the location b: it comes from the value of the above limit (1), so we must have b=1.
 limx→ b g(x) = y     ⇒ limx→ 1 g(x) = y
Finally, we need to think a little about g(x): what is its limit as x→ 1? Looking at the graph from the previous step, we see that the limit is going to come out as 1 because that is the location both "sides" are headed towards. [We could formally prove this limit to be true, but the problem does not require us to, so we can rely on a more hand-wavey explanation.]
 limx→ 1 g(x) = 1
• At this point, we have all of our ingredients prepared to create the counter-example. With f(x) = 1 and g(x) = {
 x,
 x ≠ 1
 2,
 x = 1
, we have the below limits:
 limx→ 47 f(x) = 1       and limx→ 1 g(x) = 1
The composition "law" in the problem says
 "If limx→ a f(x) =b and limx→ b g(x) =y, then limx→ a g ⎛⎝ f(x) ⎞⎠ =y."
Using our specific functions and locations, that translates to
 "Because limx→ 47 f(x) =1 and limx→ 1 g(x) =1, we have limx→ 47 g ⎛⎝ f(x) ⎞⎠ =1."

Now let us consider what limx→ 47  g( f(x) ) actually is. First, notice that since f(x) = 1 no matter what value we have for x, we can re-write the limit as
 limx→ 47 g ⎛⎝ f(x) ⎞⎠ = limx→ 47 g( 1)
We can evaluate g(1): looking at how the function was set-up, we have g(1) = 2. Thus we can once again re-write the limit, plugging in g(1):
 limx→ 47 g( 1)     = limx→ 47 2
Finally, since 2 is just a constant number, the limit as x → 47 has no effect on the value of 2: a constant is a constant. That is,
 limx→ 47 2     =     2.
Thus, by the above string of logic, we have shown that the actual limit is
 limx→ 47 g ⎛⎝ f(x) ⎞⎠ =2
We see that the actual value of the limit is different from the value purported by the supposed "law", so it must be the case that the "law" is false. [This does not mean that the opposite is always true: it just means that the "law" does not hold in every case no matter what.]
We show that the "law" fails by comparing the value the "law" gives for some limit involving g(f(x)) with the actual value of that same limit that we find by working through the limit. (See the final problem step for a detailed counter-example.) [Remarks: Although the naive version of the composition law given in the problem is incorrect, it can be fixed with a slight modification. The below version is correct and does always work:
 "If limx→ a f(x) =b, limx→ b g(x) =y, and g(x) is continuous at b, then limx→ a g ⎛⎝ f(x) ⎞⎠ =y."
Notice that the above formulation relies on the idea of g(x) being continuous at b, which is an idea we have not touched upon yet. This comes from continuity, which is a concept that is discussed in the lesson Continuity & One-Sided Limits. This also shows us an important point: our first guess at how things work is not always correct. While we might have naively expected the version of the composition law given in the problem to work, it requires a careful, critical eye to notice the flaw in that theorem. This is often the case in analytical matters. While intuition and gut feeling are important, you must temper those ideas with rigorous arguments and logic to avoid making false claims. It is only through careful proof that you can be certain your intuition is actually correct.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Formal Definition of a Limit

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Introduction 0:06
• New Greek Letters 2:42
• Delta
• Epsilon
• Sometimes Called the Epsilon-Delta Definition of a Limit
• Formal Definition of a Limit 4:22
• What does it MEAN!?!? 5:00
• The Groundwork 5:38
• Set Up the Limit
• The Function is Defined Over Some Portion of the Reals
• The Horizontal Location is the Value the Limit Will Approach
• The Vertical Location L is Where the Limit Goes To
• The Epsilon-Delta Part 7:26
• The Hard Part is the Second Part of the Definition
• Second Half of Definition
• Restrictions on the Allowed x Values
• The Epsilon-Delta Part, cont. 13:34
• Sherlock Holmes and Dr. Watson
• The Adventure of the Delta-Epsilon Limit 15:16
• Setting
• We Begin By Setting Up the Game As Follows
• The Adventure of the Delta-Epsilon, cont. 17:24
• This Game is About Limits
• What If I Try Larger?
• Technically, You Haven't Proven the Limit
• Here is the Method
• What We Should Concern Ourselves With
• Investigate the Left Sides of the Expressions
• We Can Create the Following Inequalities
• Finally…
• Nothing Like a Good Proof to Develop the Appetite
• Example 1 31:02
• Example 1, cont.
• Example 2 41:46
• Example 2, cont.

### Transcription: Formal Definition of a Limit

Hi--welcome back to Educator.com.0000

Today, we are going to talk about the formal definition of a limit.0002

In this lesson, we will explore the formal definition of a limit.0005

Previously, we defined a limit with intuitive terms, like "approach" and "goes to."0007

Now, we are getting serious, and it is time for some really technical stuff.0012

However, I want to point out that very few students will have any use for what is in this lesson, let alone during a precalculus-level course.0015

I want you to have the option to know the formal definition, if you are interested.0025

But most people will not need it, even in math class.0028

Problems based on the formal definition of a limit are extremely rare in calculus class--maybe as a bonus question, but probably no more.0031

This stuff won't come up in science classes, and it will only be necessary for high-level college math.0040

And when I say "high-level college math," I mean that this stuff is really going to only definitely start showing up0047

by the second or third year of taking advanced college math courses.0051

This stuff is not going to show up any time soon, so I really just want to be frank with you and honest:0055

what I am going to teach here is really a totally optional lesson.0060

It is not going to come up in the course that you are currently taking; I guarantee you that.0064

And it may or may not come up in the next course you take, if you go on to take calculus.0068

But it is really, really unlikely to be tested heavily.0072

It will maybe be a bonus question, or maybe one real question; but for the most part, you are not going to see this formal limit definition.0075

It just doesn't show up; it has very little direct applicability in the sciences; it is only used for some pretty advanced ideas in mathematics.0083

It is really important for building advanced ideas in mathematics.0091

But for day-to-day use of mathematics, we just never really need to see it.0094

Unless you are particularly interested in mathematics, or you know that you really want to go into high-level physics,0098

or you are interested in something like computer programming, where you will have to get enough mathematics learned,0103

at some point, that you will end up needing to be able to understand those advanced math concepts,0107

you are probably never going to end up encountering this formal definition, ever.0111

And that is totally OK; wanting to learn this sort of thing--I know that it is not for everybody.0115

But for me, I personally love this stuff; this is the reason I love math--because there are these really cool, complex,0119

strange, puzzling logic ideas that make this really interesting structure that is just fascinating--to me, at least.0127

So, I think this stuff is totally awesome.0133

If you are not interested, don't worry about it; skip this lesson.0136

There is going to be some really difficult stuff, and there is not really a whole lot of direct personal need to understand it.0139

But at the same time, there is not really a whole lot of direct personal need to go to a museum; you do it because you are interested in the art.0144

And that is the same reason that you might be interested in mathematics.0150

It is the reason I am interested in mathematics--just because I find it personally satisfying and cool.0153

If you are interested, let's go and check some stuff out.0157

OK, and before we keep going, I just want to say, for those of you who decided to stay around, thanks.0162

I think this stuff is really, really cool; I am glad to get the chance to share it with you.0166

And I think it is really, really cool; I hope you end up liking it as much as I do.0170

We few, we happy few, we mathematicians--we get to see this.0175

All right, we are going to need some new Greek letters before we get started.0180

That is right: this stuff is so cool that we need to bring out the new Greek letters.0183

Our first one is delta, and these are the lowercase versions; so δ is lowercase delta, and ε is lowercase epsilon.0187

Here is δ; and if you are going to draw it by hand, it is kind of hard to draw this thing exactly as it ends up getting typeset.0194

So, I would recommend...you want something on the top, and then sort of a curved loop like that.0201

You might end up seeing me write it like that, as I am writing it pretty quickly.0206

But that is a pretty good idea of what you are trying for.0209

This one is kind of not a very good one; but put some sort of curved loop on top, and then a sort of circular thing on the bottom.0211

And that will be enough for most people who understand mathematics to realize that you are trying to write delta.0217

You don't have to ahve it absolutely perfect, because they will know what the symbol is, especially when they end up seeing it a bunch of times.0221

Epsilon is considerably easier to draw; it is just sort of a backwards 3; that is it.0226

That is ε; that one is easier than δ, I would have to say.0231

All right, the formal definition that we are about to see involves these letters.0233

So, it is sometimes called the (ε,δ)-definition of a limit, or just the formal definition of a limit.0238

We just talked about the intuitive idea of what a limit is; and that is a really great way to think about limits.0244

But sometimes you really need to get to the basic facts and turn intuitive ideas into precise things, so that you can say "this is what it is."0249

And that is what this definition is for.0256

All right, let's get to it: the formal definition of a limit.0257

Let f be a function defined on some interval a to b of the real numbers, where a is less than b.0261

Let c be contained in a to b, and let l be a real number.0268

Then, the limit, as x goes to c, of f(x) = l means that, for any real ε greater than 0,0274

there exists some δ greater than 0 such that, for all x where 0 is less than the absolute value of x - c,0282

which is less than δ, we have the absolute value of f(x) minus l is less than ε.0289

What? Don't worry: if you found that definition confusing, that is exactly how I felt when I first heard it.0296

The first time you see this, you think, "What is going on?"0304

That is totally OK; it is something that you pick up and work through slowly, over time.0307

If you opened a book for the very first time, you wouldn't expect to be able to understand every word on the page.0311

It is something you have to work through; it is OK--you are building up these ideas of understanding it.0316

It is not something that is going to make sense exactly the first time.0319

It is something that makes sense over time; that is totally fine.0322

So, we will start working through it, step-by-step, and we will see some ways to understand what is going on in this definition--0325

to see how we can parse this thing and make sense of it.0330

All right, first, let's get the basics out of the way, the ground work here.0333

The first half of the definition is just setting up the limit; that is what we end up getting in this first half.0337

Let f be a function defined on some interval a to b (they are real numbers) where a is less than b, and c is contained in a to b, and l is some real number.0342

All right, let's break that down, piece by piece.0352

First, let f be a function defined on some interval a to b.0354

That just means that it is a function, and it is defined over some portion of the reals.0358

There is some piece of it that it ends up working in.0363

It is basically just saying that this function is defined somewhere; that is all that we are saying.0366

There is some chunk of the reals where this function makes sense.0370

This part where we say a < b is just a guarantee that we don't accidentally say a = b, and we end up having something that doesn't make sense.0373

It is so that we know that we are going from some a, up until some b; that is our interval of where the function is defined.0379

Next, the horizontal location c, contained inside of (a,b): remember, this symbol just means "inside of."0386

So, (a,b) is a set of all possible values; it is the interval, and that symbol just means that c is inside of that set of possible values.0393

So, the horizontal location c contained inside of (a,b) is the value that the limit will approach.0400

That is the thing where we are getting closer and closer to that, and we are seeing where we end up going to.0406

So, c is just a location that we are getting towards; that is the value that the limit will approach.0412

And finally, l is some real number; l's real number is the vertical location that we end up going to.0417

That is where the limit will go to; we are setting the l that we are saying our limit ends up going to as we approach this c.0424

So, that is the basics that we are setting up: f is some function; it works on some chunk.0432

c is where we are going to, and l is the value that we end up meeting up at, that we are headed towards.0436

All right, the hard part is the second half of the definition.0443

The limit as x goes to c of f(x) = l means that, for any real ε greater than 0, there exists some real δ greater than 0,0447

such that for all x where 0 < |x - c| < δ, we have f(x) - l < ε; OK.0455

To break this up, we want to be able to focus on this one piece at a time.0462

So, what we are going to do is focus on the ε portion first.0465

It will be easier to understand that; and then we will see the δ part in that idea.0468

So, for any real ε greater than 0: what that means is that, if we just grab some ε--0473

we just say some number, like 1 or 1/2 or whatever number we feel like saying,0479

we will end up having this be true later on: that the absolute value of f(x) - l is less than ε.0483

Now, that might be kind of hard to understand: the absolute value of f(x) - l is less than ε...0490

we are not used to working with absolute value that much; it is kind of confusing.0493

But remember one of the things we talked about when we first talked about absolute values.0496

Absolute value is a way of seeing the distance between two things.0499

If we have the absolute value of m - n, if we have the absolute value of q - r, that is a way of saying the distance between those two points.0502

The absolute value of some object minus some other object is just a way of saying how far those two objects are from each other.0512

That is one of the ideas that we have from absolute value.0519

When we say the absolute value of f(x) - l, what we are really saying is how far f(x) and l are from each other.0522

Now, that means, with the combination of ε being greater than 0, and the distance between f(x) and l being less than ε...0529

what we are doing is saying that there is this boundary around l.0537

The maximum distance f(x) can be from l, the distance between f(x) and l, has to be less than ε, this thing right here.0540

Since the distance between our f(x) and our l is less than ε, it means that our f(x) has to be within ε distance of l.0549

We effectively create a boundary that we hold f(x) inside of.0556

Whatever our f(x) ends up being, it has to be somewhere inside of this boundary.0560

If it is outside of the boundary, like this, that is not allowed, because it would fail to be within ε distance of the l.0565

We end up seeing...we have that our l is here, and then the ε's here are just the distance that we end up having for that boundary.0574

We are setting a boundary that is ε distance out from the l.0583

But notice: it has to be true for any ε that we set, so we will be able to create a boundary at whatever size we decide to make,0587

whether it is a giant ε boundary or a really tiny ε boundary.0593

It will always end up working for this next part that we are about to work towards.0596

OK, the second half of the definition (continuing with that): previously, we saw ε as a boundary around the height l.0600

The absolute value of f(x) - l is less than ε means that f(x) has to be, at most (has to be less than), the distance of ε from l.0607

We know that we have to be somewhere contained inside of this boundary in here.0617

OK, now let's move on to looking at δ.0624

What does δ get us? What we did is went and set up for any real ε greater than 0; we had this boundary get set up.0628

So, with that boundary in mind, we now have to go on to say that there exists some real δ greater than 0,0639

such that, for all x where 0 is less than x - c, which is less than δ, we will end up having this |f(x) - l| < ε be true.0646

So, what we are doing is putting a restriction on the allowed x-values.0657

The δ sets some restriction on it: you can't get any farther from c than δ.0663

So, what we are doing is saying, "Remember: the |x - c| is saying the distance between x and c."0669

So, the distance between x and c has to be less than δ.0677

So, if the distance between x and c is less than δ, that means that we have set up, once again, a boundary of how far our x is allowed to roam.0681

And we know that what has to come out of this is: we have to have...0689

if we set x's within that boundary, we will end up having the |f(x) - l| being less than ε.0693

So, if you use x's from inside of this boundary here, we know that we end up having to be mapped into the f(x) that is within that vertical boundary.0702

We are setting up some set of boundaries on what x is that we can have.0713

And we know for sure that we have to have it so that they end up getting mapped inside of this ε.0718

The other thing to notice is that it says that 0 is less than |x - c|, that 0 is strictly less than the absolute value of x - c.0724

It means that x cannot be equal to c, because then we would have an absolute value of 0.0731

We would have x = c; but since it is a limit, we are not concerned with x = c.0735

Remember: we are not concerned with the destination; we are concerned with the journey.0738

So, that means we care about the part with the boundary, but not the actual thing in the middle.0741

That is what that 0 < |x - c| means: that we are guaranteed that we can't have x equal to c,0745

and |x - c| < δ means that we are bounded δ distance, at maximum, from the c in that middle.0753

And everything inside of it ends up getting mapped inside of this boundary here.0761

It will be easier to see what is going on if we erase some of this.0767

What we are doing here is: we can think of it as: some ε gets created around our l that sets up a boundary.0770

We know that you have to be somewhere on this portion vertically.0781

And then, we set some δ radius around our c, so that it ends up mapping only to those vertical locations there.0785

We set ε, and then we have to be able to always set some δ from that ε0797

that will end up mapping us inside of that vertical chunk--that we say that,0801

if we are close enough to this value of c, we will always come out close enough to that value of l.0805

OK, now the really important idea is that we can do this for any ε greater than 0 whatsoever.0813

That is, for any ε greater than 0, there has to exist some δ greater than 0,0818

such that, for x within δ of c, if we are close enough to c, f(x) will end up getting mapped within ε of l.0825

If you are close enough to c, you will end up being close enough to l.0833

We can imagine this process as a never-ending dialogue of ε challenges against δ defenses.0836

Someone says, "Is it possible to stay within this distance of our l?" and we say, "Yes, as long as we are within this distance of our c."0842

And then, they say, "OK, is it possible to stay within this distance of our l, this ε of our l?"0849

We say, "Yes, as long as we are within δ distance of c."0853

So, some ε gets named; for any ε named, there has to be a δ that will cause us to stay within ε away from that l.0856

Otherwise, if we can't do this, then the limit does not exist.0865

But if we can do it, forever and ever and ever, then that means that the limit does exist.0869

If for any ε that gets named, there is always a δ, then that means that the limit does exist.0872

Now, of course, we can't actually have this process of ε challenge and δ defense; we can't do this forever.0877

We can't do this for eternity, so we have to figure out another way to do this.0882

If we wish to truly prove a limit, we have to prove that there will always be some δ greater than 0,0885

given some ε greater than 0--that if some ε gets named, there is always a method0892

that will create the appropriate δ, so that we can get within ε of that limit l.0896

OK, now what we are going to go onto is: we are going to see both of these ideas.0902

Both of these ideas will be explored in the next scene that will be played out by Sherlock Holmes and Dr. Watson.0906

All right, the setting is late afternoon in Sherlock Holmes's apartment.0912

He and Dr. Watson are lounging about.0918

"Let us play a game, Watson!" "All right, Holmes, what type of game?"0921

"A game of limits." "I do not believe I have ever..."0924

"Do not worry: I shall teach it to you. We begin by setting up the game as follows."0927

"One of us (I, in this case) claim the existence of a limit; I name some function f(x)0932

and some value l that the function will approach as x approaches some number c."0938

"Written out, we have the limit as x goes to c of f(x) is equal to l."0944

"All right, that makes sense enough." "Good; after that, the game in earnest begins."0951

"The opposing player (you, in this case) names an ε greater than 0."0955

"It will then be the first player's job to name some δ greater than 0, such that,0959

for all values of x within δ distance of c, except x = c (we don't care about x = c in this game),0964

they will cause f(x) to be within ε of l."0972

"So, if we use one of those x's that is within δ of c, we know that the f(x) that will be produced by that x will have to be within ε of l."0975

"I think I understand." "Let me write it out: I will name some ε greater than 0 that sets up some boundary around l."0983

"So, we have some l - ε to l + ε boundary; we are going to be in that interval vertically."0990

"Then, you name a δ greater than 0 such that, for x contained within c - δ to c + δ,0997

that horizontal bounding, we will end up getting f(x) contained in l - ε to l + ε;1005

that is to say, the boundary that we set around l in the first place."1012

"I will set some boundary vertically, and then you will respond with some horizontal boundary that will put our function inside of that vertical boundary."1016

"Very good, except one important thing: we do not concern ourselves with x = c."1024

"For this game, x is not equal to c; we never actually consider x equaling c, so x is not equal to c, generally."1034

"All right, I understand: you set up a limit; I challenge it with some ε greater than 0, and then you defend with some δ greater than 0."1042

"One question, though: why can't ε equal 0?"1050

"I understand that δ must be greater than 0, because we must have x not equal to c;1054

if δ is greater than 0, then we know that we can't actually be on top of x = c, so x is not equal to c. But why not ε = 0?"1059

"Because this is a game about limits! If you set ε = 0, we would be stuck on l."1067

"We wouldn't have a boundary around it; we would be stuck there."1074

"We need a boundary around l, so that we can talk about approaching it;1077

the entire idea about a limit is to talk about going towards it, approaching it."1082

"It is similar to why we must have x not equal to c; otherwise there is no approaching it, if we are stuck on top of it."1086

"So, we can't be stuck on top of it, either vertically or horizontally; that is why we can't have ε or δ equal to 0."1092

"Makes sense; so what comes next?" "That is it; we stay with the same limit,1098

but the process of ε challenges and δ defenses repeats for as long as we care to play."1103

"If you manage to give an ε that stumps me, you win! You will have shown the limit false."1109

"Let us begin; what will your limit be?" "I choose the following limit to defend: limit as x goes to 3 of x2 + 2 = 11."1115

"All right, I begin by challenging with ε = 1." "I defend with δ = 0.1."1123

"Since x contained within the interval 2.9 to 3.1 (notice: that is a δ of .1 away from our c of c = 3),1130

if x is contained within 2.9 to 3.1, it will give f(x) contained within 10.41 to 11.61."1141

"And that means I am easily within an ε value of 1 from our l of 11."1148

"Thus, my defense stands: using δ = .1, I stay within ε of 11."1153

"OK, how about ε = 0.05." "I will go with δ = 0.005."1159

"Since x contained within 2.995 to 3.005 causes f(x) to be contained within 10.97 to 11.03, I am within ε of 11, so my defense stands again."1167

"What if I tried larger...I set ε = 10!" "Nice try, old chap, but I already showed that δ = 0.1 works for ε = 1."1179

"Therefore, it must work for ε = 10 as well; if you started out small, and then you get bigger,1190

well, I can just leave my δ the same, and it will work just fine for the larger vertical boundary, as well."1195

"Thus, there is no hope for going larger with your ε.1200

Since I have already shown that it works for a small ε, it will automatically work for a larger ε, as well."1204

"Hmm...fine, deal with this: ε = 0.000001." "Elementary, my dear Watson: δ = 0.0000001."1209

"Notice that, if x is contained within 2.9999999 to 3.0000001, we will have f(x) contained within 10.9999994 to 11.0000006."1220

"That is to say, it will be contained within an ε of 11; thus, my defense is still standing."1237

"All right, Holmes; you have me convinced--I believe that, no matter what ε I say, you will be able to defend with an appropriate δ."1244

"But technically, you haven't proven the limit; we can't play this game for eternity."1251

"So, how can you prove that δ will always exist?"1257

"An astute question, Watson; the answer is by showing that there exists some method1261

to create δ from any given ε, and that the method never fails."1266

"I will tell you the method I am using, and then show that it always works."1272

"Of course, you already have a method." "Naturally; here it is: for any ε greater than or equal to 1,1276

I will simply respond with δ = 0.1; as we discussed earlier, δ = 0.1 works for ε = 1,1281

so it will work for any ε greater than or equal to 1."1288

"On the other hand, for ε between 0 and 1 (that is, 0 < ε < 1), I will use δ = ε/10."1291

"Wait; how do we know that δ is greater than 0?"1301

"Well, we know that ε is greater than 0; and since we are creating δ1304

by dividing ε by 10, we know that δ must be greater than 0, as well."1310

"If ε is greater than 0, then we know that dividing something that is greater than 0 doesn't cause it to become 0,1316

and doesn't cause it to become negative; it simply makes it smaller."1322

"So we know that δ must be greater than 0, as well, because we simply divided1325

something that was already greater than 0 by some positive number--in this case, positive 10."1329

"Now, let me prove to you that δ = ε/10 always works."1333

"Notice that, since we are working with the function x2 + 2,1339

the values farthest from c = 3 will produce the values farthest from l = 11, because 32 + 2 = 11."1343

"Thus, we only have to concern ourselves with x = 3 - δ and x = 3 + δ.1352

"If that didn't quite make sense, Watson, what we have here is that, at some horizontal location 3,1356

we know that it puts out the vertical height of 11."1362

"So, as we get farther and farther to either side, we will end up getting farther and farther to 11."1365

"We know that this is a parabola; we are used to working with parabolas; so we see that, the farther we end up getting away from c = 3,1371

the more extreme we are from c = 3, the more extreme we will be from our l = 11."1378

"That is how 32 + 2 would work; as long as we don't go so incredibly far1384

that we end up wrapping the parabola to the other side, we are safe from this; we don't have to worry about it."1388

"So, as long as we can be absolutely sure of that fact, where we don't to -3 or go past 0, we will be fine."1393

"And we know that δ must be less than or equal to .1, because of our earlier restriction on it always being less than or equal to .1."1400

"So, we know that we can't get too far; the most extreme values for our x will end up being using the whole of δ--1407

that is, we only have to concern ourselves with x = 3 - δ and x = 3 + δ."1414

"Now, I want to point out, before we continue, that this logic is specific to the function we are working with, Watson."1420

"It is specific to working with x2 + 2, the fact that, as we get farther and farther away horizontally,1425

we will get farther and farther away vertically, as well; so looking at maximum horizontal distance means maximum vertical distance."1431

"A different function, though, might require different logic; so we have to think specifically1438

about the function that we are working with, if we are going to prove something."1442

"In this case, though, we see that this method will work."1445

"Thus, we can show that our δ = ε/10 works by showing that, if we use the above x1448

(that is x = 3 - δ and x = 3 + δ--if they both work) in our function,1454

for being within ε of l, if we have that they end up coming out like that,1460

we know without a doubt that our δ will always work, because it worked for the most extreme possible values."1467

"We must show that the absolute value of 11 minus f(x)...in this case, if we are plugging in 3 - δ,1474

and our f(x) is x2 + 2, we would have the absolute value of 11, our l, minus...1481

plugging in our function's maximum, most extreme possible value for x, that is (3 - δ)2 + 2 must be less than ε."1486

"Similarly, over here, for our other extreme value for x, x = 3 + δ,1495

if we plug in our f(x), we have the l that we are going to (11) minus the quantity (3 + δ)2 + 2, is less than ε."1501

"Those are the most extreme values there; we are doing the absolute value of l minus f(x) must be less than ε."1511

"That is what we want to show; so to show that this is the case, we will investigate the left sides of the expressions."1519

"Let's work with the 3 - δ part first; we have |11 - (3 - δ)2 + 2|."1524

"Let's begin by expanding 3 minus δ squared; we expand 3 minus δ squared, and now,1531

we have the absolute value of 11 minus (9 - 6δ + δ2); 3 - δ, squared, gets us 9 - 6δ + δ2."1535

"We have minus here, and we have 9 and 2 inside; so minus 9, minus 2, cancels out the 11."1544

"We are left with -6δ + δ2, but there is still this minus sign, so it distributes on,1551

canceling that and making this negative; and we have the absolute value of 6δ minus δ2, when we plug in x = 3 - δ."1560

"Over here, on the other one, we have |11 - (3 + δ)2 + 2|."1569

"We expand 3 + δ, squared; we get 9 + 6δ + δ2;1576

so now we have |11 - (9 + 6δ + δ2) + 2|; once again, we are subtracting by what is inside of that."1581

"So, the 9 and the 2 go together to combine and take out the 11, leaving us with 6δ + δ2."1588

"The minus sign distributes, and we now have -6δ - δ2 if we had plugged in x = 3 + δ."1593

"So, that is what the distance between l and our f(x) is when we plug in our most extreme possible x's, 3 - δ and 3 + δ."1602

"Continuing from this, we can see that, since 0 is less than δ, which is less than or equal to 0.1--1611

remember: we set that δ must be less than or equal to 0.1 at the beginning,1616

because the largest ε that we were allowed to really care about was ε = 1,1619

and we know that δ = 0.1 works there, so we always kept our δ's smaller than 0.1, and δ must be greater than 0."1624

"We can write the above as 6δ - δ2 and 6δ + δ2."1631

"If you are not quite sure about this, since 6δ...well, we know that δ must be greater than 0;1636

thus, 6δ must be positive, and minus δ2...well, δ2 must be smaller than δ,1641

because δ is less than or equal to .1; if we square something that is smaller than 1,1648

it must end up being smaller than had it not been squared."1655

"So, 6δ - δ2...well, 6δ is positive; minus δ2 is negative, but smaller than 6δ."1659

"So, when we take the absolute value of this expression, we are left simply with 6δ - δ2, even after the absolute value."1666

"Over here, we have -6δ - δ2; well, that means the -6δ and the -δ2 are both negative."1673

"So, they combine forces and, after the absolute value is finished with them, they will end up coming out as positive, so we will have 6δ + δ2."1679

"From this, we can create the following inequalities, the right side, in this case,1687

being because δ < 0.1 implies that δ2 < δ, which we talked about very recently."1690

"If we have 6δ - δ2, well, if we had 7 - 5, that would clearly be less than simply 7, if we got rid of the thing subtracting from it."1696

"So, we just get rid of the thing subtracting from it, and we have 6δ - δ2 < 6δ."1706

"That is certainly true; furthermore, since δ2 is less than δ,1712

we have that 6δ + δ2 < 6δ + δ, which is the same thing as 7δ."1716

"So, we now know that 6δ + δ2 is less than 7δ."1723

"At this point, we can use our δ = ε/10 that we originally set; and we have 6δ = 6ε/10."1727

"Remember: that is less than the most extreme value; and 6ε/10 is clearly less than ε."1736

"Furthermore, the other most extreme value was going to be less than 7δ, and we have 7ε/10,1741

which is indeed less than ε; thus, we have shown that plugging in our most extreme values1746

will show that the distance between f(x) and l is always less than ε."1751

"We plug in our most extreme possible value; that gives us our most extreme possible vertical distance,1756

and it still ends up being less than ε when we work the whole thing through."1761

"So, the method of choosing δ always works; we see that this δ = ε/10 will always work; it will never fail."1765

"Very good, Holmes; my only question is how you decided on setting δ equal to ε/10 in the first place."1774

"How did you think, 'ε/10; that is the thing for me'?"1782

"Indeed, that is the most difficult part of proving a limit."1785

"In general, once you have a sense of how the function works, it helps to set it up as if you knew what δ was."1789

"As you work forward, you will find its requirements."1797

"For example, with this limit, I realized that whatever δ was, whatever I ended up choosing for δ,1800

it would end up having to get to the point where we would see that 6δ + δ2 had to be less than ε."1805

"So, if 6δ + δ2 had to be less than ε, I could restrict δ < 0.1,1812

because a smaller δ will never cause any harm, and see that 6δ + δ21817

would be the same thing as saying that it is less than 7δ; and 7δ would certainly work if δ was ε/10."1823

"I could have gone with ε/7, but I figured, 'Why not have a little bit of extra room?'"1830

"And ε divided by 10 just seems so nice and round;1834

so I saw that δ = ε/10 would do fine, at which point I was prepared to work through the proof."1836

"Ah, I am feeling a mite peckish; nothing like a good proof to develop the appetite--what are you in the mood for, Watson?"1842

"Somehow, I find myself craving Greek." "Excellent--good! A brace of gyros it will be!"1848

And they make their exit for the neighborhood gyro shop.1854

All right, with Sherlock Holmes and Watson having helped us see what is going on here,1857

the idea of an ε challenge and a δ defense, and how to prove this stuff in general,1860

by showing that this δ method will always give us a δ that works, we are ready to start working through some examples.1864

Our first example: Below is the graph of f(x) = xsin(1/x).1870

First, we need to determine if the function has a limit, as x goes to 0, and if so, find it and prove it.1876

We can see, right from the beginning, that it looks like it is going to 0.1883

We can see from this nice, convenient graph that it looks like it is going to end up having a limit.1887

It would appear to be the case; however, we notice that we can't simply plug in x = 0.1893

If we did that, we would have 0sin(1/0); we are dividing by 0, so f(0) does not exist.1900

We can't simply go down that method; we can't simply plug in a number and figure out what is going to come out.1910

So, we have to show this by proving it directly; we have to go and see that this is going to always work.1915

We have to figure out a way; so we can see from the graph that, yes, it makes sense that it is going to end up coming out to be 0.1921

But it will be our job to now prove it through a good proof.1926

Let's try to get an understanding of why this thing looks like this.1931

We can see, from the graph, that it looks like this; but let's see what is going on.1934

What we can do is break this down into multiple pieces.1937

We can first see this as sine of...well, let's look, actually, first, at 1/x; what does 1/x graph as?1940

Well, 1/x is going to end up graphing something like this; we are going to go asymptotic as we approach that vertical axis.1951

Then, how does sine of some value graph?1960

Well, sine of some value ends up graphing like this; it has that nice repeating nature.1965

If you watched the last lesson, you probably ended up seeing this in the previous lesson.1975

We talked about just sin(1/x), and how it went crazy and came really, really close.1979

That is the same idea so far; we have that part going on so far.1983

If we plug these two things in--if we plug 1/x in for sine of t, then what we are concerned with,1987

as we get close to 0, as x goes to 0, is that part that is getting close to that.1992

If we end up looking at that, if we look at sin(1/x), we are going to end up getting this part where,1998

as it is far away from 0, it is going to end up going sort of slowly.2008

But as it gets closer, it is going to speed up, and it is going to start to go crazy.2011

And it is going to bounce back and forth between 1 and -1 really, really fast, because it has to go through all of infinity.2015

We shoot off to infinity as we get towards 0; so since it is shooting off towards infinity, 1/x is shooting off towards infinity inside of the sine.2021

So, the sine is now going to have to speed up faster and faster and faster, because it has to go through all of infinity by the time it makes it to 0.2029

We see the same thing from both sides: it shoots up and gets really, really, really, really fast, bouncing up and down.2035

And so, that is the behavior we see.2040

If it was simply sin(1/x), it wouldn't have a limit, as we talked about in the previous lesson,2041

because it is never going to agree on a single location that it is working towards.2046

However, in this case, we also have x showing up; well, if we graph just x, what does just x look like?2051

Well, that looks just like this; so x times something is going to end up expanding it or shrinking it by whatever x's current value is.2059

As x gets closer and closer to 0, it is going to end up squishing it and squishing it and squishing it, closer and closer to that x-axis.2070

Now, notice: sine of anything always produces values that are going to be between -1 and +1,2076

based on how the unit circle works and what we can see from the graph of just sine.2089

So, what we are going to have is either a positive x, effectively, or a negative x.2095

So, if we graph in that -x...here is -x in green...then as we plug in x times sin(1/x), it is going to get...2099

the maximum value that they can have is whatever the x is that it is next to.2114

So, as it gets far away, it ends up bouncing between these things.2117

But as it gets closer and closer and closer, it ends up bouncing faster and faster and faster and faster.2120

But because there is that x, it shrinks it down and crushes it to go down to 0.2125

So, it is not actually going to be defined at x = 0; but it will end up being crushed into this point right here.2130

So, with all of these ideas in mind, we can now say that the limit as x goes to 0 of x times sin(1/x) equals 0.2137

We figured out that, yes, it does have a limit; it makes sense; we understand what is going on2150

in a thing that now makes sense with this graph.2155

And notice: you can also see that the x is going out like this in the graph.2157

And then, -x goes out here; so this nice, computer-generated graph--we end up seeing the same behavior2165

of the sine bouncing infinitely quickly...it gets trapped between +x and -x, because it has to multiply x times sin(1/x).2172

So now, we see that this is what our limit ends up being.2181

And at this point, we can try to work out a proof here.2184

So, if we are going to prove that this is the case, then it must be that, for any ε greater than 0,2187

there exists some δ that will end up making us within that appropriate distance of our vertical ε.2195

For any ε greater than 0, what δ do we choose--what is the appropriate method?2204

Well, notice: if we go back to our x (here is a graph of x, once again; and I will also map the -x here, as well...2208

sorry; that should be dots all the way through); if we have -x and x mapped here, well, what is the biggest thing that x allows?2221

Well, the biggest thing that x is going to allow that sine to get expanded out to is whatever x currently is.2229

For any ε greater than 0, if we want to talk about the maximum vertical size we can allow,2235

then we can allow x to go out as far as that ε, and that will be the maximum allowed vertical size.2238

So, we can simply say that our δ (since that is how far out our x can go from 0) is simply equal to ε.2244

We set δ = ε now, it is going to be our job...what we want to show now is that,2251

for any f(x) - l, it is going to be less than ε if we have our x within δ distance of our c.2260

All right, so let's start working through this.2276

We can swap out what f is and what l is, and we can swap things out, and we can start working and showing that this is true.2277

f(x) - l...what is the left side there? f(x) is this thing right here, x times sin(1/x), minus l...2286

well, l is just 0 in this case, so minus 0...it just closes up, and we have that it is less than ε.2297

So, at this point, we can look at this and say, "Well, x times sin(1/x)...the absolute value of all of that2304

is just going to be the same thing as the absolute value of x, times the absolute value of sin(1/x)."2308

If we strip any possible negative signs after they have multiplied,2313

or strip any possible negative signs before they multiply, it is not going to have an effect on the product.2316

So, we have that this thing right here...2320

Oh, we don't know that it is less than ε; what we want to show is that it comes out to be less than ε.2321

We want to work towards showing that; I'm sorry about that.2326

The absolute value of xsin(1/x) is just the same thing as |x| times |sin(1/x)|.2328

Well, what is the absolute value of sin(1/x)?2338

Well, the absolute value of sin(1/x)...the absolute value of sine of anything at all is going to be less than or equal to 1.2341

You can't get larger than 1 or larger than -1 when you plug something into sine.2355

The absolute value of sin(1/x) is going to still be less than or equal to 1,2360

except for that specific value of x = 0, which would cause the whole thing to break.2365

But we don't have to worry about that, because it is x going to 0; so we have 0 < x - c.2369

So, we don't have to worry about x = 0; so we know that sin(1/x) ≤ 1 will always be true,2375

because we don't have to worry about actually plugging in x = 0.2380

So, we can now use this information up here; and we have that |x| times |sin(1/x)| is less than or equal to...2383

we swap out that fact that sin(1/x) is less than or equal to 1, and we have |x| times 1.2391

So, |x| times |sin(1/x)| is less than or equal to |x| times simply 1.2397

We have that that is just the same thing as saying x; great.2403

Now, at this point, we also know that x - c is less than δ.2407

Let's use yet another color here; so...well, what is our c?2412

Our c is equal to 0 in this case, because that is what we are going towards.2416

So, we have that the absolute value of x - 0...well, we will just leave it as the absolute value of x...has to be less than δ.2420

The absolute value of x is less than δ; we can now swap that in here.2427

And we have that the absolute value of x is less than δ.2431

What did we set our δ to? We know that δ is equal to ε, because that is what we decided to make our method.2437

We have that that is equal to ε; so at this point, we have now shown that the absolute value of x,2442

times the sine of 1/x, gets to being less than or equal to x, which we know is less than δ, which...2450

δ is equal to ε, but at this point we can stack together our signs.2459

The most extreme sign that we ended up having was that direct strictly less than;2462

so at this point, we now see that the absolute value of x times sin(1/x) is, indeed (close that absolute value), less than ε.2466

So, that always ends up being true, if we use this method.2478

We have now completed our proof; this limit does, indeed, work out.2481

We have shown that whatever ε we end up choosing, there will always be an appropriate δ,2485

if we simply set δ = ε it will end up satisfying any and every ε.2490

Thus, we have completed the proof; the limit does exist; the limit is, indeed, 0.2496

Awesome; all right, the second example (and there are only going to be two of these examples,2500

total, because they clearly take a little while to get through):2503

consider the piecewise function f(x) = 1 when x is rational, -1 when x is irrational.2506

Prove that f(x) has no limit anywhere.2512

The very first thing we want to do here, before we try to prove anything, is understand just what f(x) = 1 when x is rational, -1 when x is irrational, means.2515

First, we need to remember what it means for a number to be rational.2524

A number is rational when it can be put as some a/b, where a and b are integers; that is what it means for a number to be rational.2528

A number is irrational when it can't be expressed as a rational.2540

A rational means any decimal number that has a fixed length to it, and also decimals that end up having repeating patterns.2544

But it will actually be enough, with just a fixed length, to understand what is going on here.2551

2.000000001, stop, is a rational number.2555

An irrational number is a decimal number, or something that can be expressed as a decimal number,2562

where the decimals continue going on forever; they never establish a single pattern,2566

and they just constantly are changing forever and ever and ever.2570

For example, π is 3.1415...stuff going on forever; √2, e...we have seen some irrational numbers that are pretty important.2572

But they are also everywhere total; imagine...we had that rational at 2.000000001,2584

but we could also have an irrational that is right next to it, practically, at 2.000000001...2590

and then a random string forever and ever, like 158234671111583...it just keeps changing and changing and changing.2595

So, we can have an irrational number right next to any rational number.2608

We can get as close as we want to any rational number.2612

And we can get as close as we want to any irrational with a rational, by just putting as many decimals as we want, and then stopping at a certain point.2615

So, the rationals and the irrationals are right next to each other, everywhere total.2621

Every single place on the number line has a rational and then an irrational, effectively, infinitely close to it;2627

and then, a rational right next to that and an irrational right next to that...everything is just infinitely mashed together.2632

2 isn't both a rational and an irrational; 2 is simply rational.2638

But right next to it is an irrational; you can get as close to it as you want, with 2.0000000...random pattern.2642

And then, you can get as close as you want to 2.000000...random pattern with a rational, like 2.0000000000000, stop.2648

In matching the random pattern 2.00000000....match the first 50 thousand digits of the random pattern,2658

and then stop, and that is still going to be a rational number, because it is technically an integer divided by an integer.2664

It will be a large integer divided by a large power of 10, but it is still technically an integer over an integer, so it is going to be a rational number.2668

What that means is that rationals and irrationals are right next to each other everywhere.2676

If we tried to graph this, what would f(x) end up looking like?2682

Well, we would end up having some that looked like this, where, let's say, we use blue when one x is rational;2685

so we are going to end up having...everything that ends up being a rational will come out at 1.2693

There are always going to be points, but between any point and the next point over,2700

it is going to end up having a little, tiny hole for the irrationals.2703

And where do the irrationals show up? Well, they show up at -1.2707

So, we end up having the same thing here; down below them are our irrationals.2709

And so, there are a bunch of irrationals; there are infinitely many irrationals down there; there are infinitely many rationals up here.2714

And they both are just right next to each other.2720

You are constantly jumping back and forth and back and forth and back and forth and back and forth, if we were to try to plot this thing out.2722

It is certainly not continuous, because it is two things simultaneously.2728

So, why can't we have a limit? Now let's try to think about why there will be no limit anywhere.2731

Well, consider: if we tried to talk about some location c, then that means that that location c would have to be in either the top or the bottom.2737

And if it ended up being in either the top or the bottom--let's just say the top for argument's sake--2746

then that means that we can put some ε boundary that is small enough to contain just one of these.2750

We make ε equal to a half or something, so that it can just contain one of these height strings, one of these lines.2757

So, since it can only contain one of these lines, no matter what δ we end up choosing,2765

if we choose any δ, then we are going to end up having points that are both on the bottom and on the top.2770

So, if we were in the top, we can make ε only contain the top;2779

but whatever δ size we choose, since the rationals and the irrationals are right next to each other,2783

if we have some rationals, we have to have some irrationals with it, as well.2787

If we have some irrationals, we have to have some rationals with it, as well.2791

If you catch this stuff, it has both types, no matter what, if you take any interval chunk of it.2794

So, since δ will always produce some interval chunk, it has to have both rationals and irrationals in that interval chunk.2800

We grab a chunk, and it is going to have both the top and the bottom,2806

even though we have restricted our ε to only contain either the top or the bottom.2810

And this is the idea of why there can be no limit anywhere: because we can always choose an ε2814

that is going to only take either the top or the bottom line, but no matter what δ we end up choosing,2820

we are going to have to be in both the top and the bottom lines.2825

So, we won't be stuck just inside of that ε it will fail to end up having a limit.2827

If we are going to actually prove this formally, that will be a little bit difficult.2835

But understanding the idea is 90% of the battle.2839

I would much prefer that you understand this idea, and the proof doesn't make sense,2841

than that you really, really work hard through the proof and have a vague understanding of doing the proof,2846

but you don't understand the idea of the limit.2849

At this point (and probably ever), the most important thing, by far, is understanding the general idea,2851

and having a sense of what is going on, rather than being able to do the specific mechanics.2858

The specific mechanics can always come later; understanding the idea is the goal.2862

All right, but let's work through those specific mechanics.2865

OK, proof by contradiction; how do we show this?2868

We will show by assuming it is possible; and then we will see that, if we were to assume that it is possible, crazily impossible things will occur--2871

that if this were true, that we did have a limit, then it won't work--it just doesn't make any sense.2879

And so, we will have contradicted it, and we will know that it must be the case that the limit cannot exist.2885

So, what we do is begin by assuming (there will be a lot of writing for this; I'm sorry)2890

that there exists some limit that as x goes to c; our f(x) is equal to some value of l.2900

Then, for any ε greater than 0, by the definition of how a limit works,2909

we know that there exists some δ greater than 0 such that...2917

and from here on, when I write "such that," I will probably just use s.t. if I have to write that again...2925

such that, for any 0 < |x - c| < δ--that is to say, an x that is less than δ distance2930

away from whatever our c happens to be--we know that it must be the case that the absolute value of f(x) - l,2943

that is, the distance between the f(x) that gets mapped from that x and l, what our limit goes to--has to be less than ε.2951

So, that has to be true; that is what it means for it to have a limit.2959

If there exists a limit, then it must be that for any ε greater than 0, there will always exist a δ greater than 0,2964

such that if x is within δ of c, f(x) must be within ε of l; that is what the formal definition is.2969

Now, let us consider (let me actually change colors, because now we are changing ideas)...2978

that was the setup to this thing: the next thing is: Consider ε = 1/2.2984

Therefore, there must exist, by this thing up here (for any ε greater than 0, there exists a δ greater than 0)...2995

there exists a δ greater than 0 such that it does this thing here.3003

There exists δ greater than 0, such that it follows this idea right here.3010

Notice: by the absolute value of f(x) - l being less than ε, being less than 1/2, it must be either...3022

well, what are the possible values that can come out of f(x)? 1 and -1.3042

It must be either |1 - l| < 1/2, or |-1 - l| < 1/2.3047

The distance that our f(x) is from l must be less than 1/2; so either 1 - l is less than 1/2, or -1 - l is less than 1/2.3058

That is what |f(x) - l| < 1/2 has to come out meaning.3066

But also notice: only one of these can be true.3071

These two ideas right here, |1 - l| < 1/2 and |-1 - l| < 1/2, can't both be true,3085

because if our l is within 1/2 of -1, it can't possibly also reach to being 1/2 of 1.3094

And if our l is within 1/2 of 1, it can't possibly also be within 1/2 of -1.3101

It has to pick one of the two bands; our l can't be 1/2 away from both bands, because 1 and -1 are two apart.3107

If you are within one of them, you can't be near both of them; so it can't be near both of them.3115

Only one of these can be true; we are going to use that fact very soon.3123

Now, we go on to say, "Well, there existed some δ greater than 0 such that it made 0 < |x - c| < δ, which means that |f(x) - l| < ε."3129

So, by our δ, we know that 0 is less than x - c, which is less than δ,3141

and that that ends up having to mean that the absolute value of f(x) - l has to be less than our 1/2, this thing right here.3155

We know that our δ has to end up making our f(x) within 1/2 of l, since we set our ε as 1/2.3166

But 0 < |x - c| < δ--that is to say, x is within δ of c--must contain rational and irrational x.3173

Since we know that δ has to be greater than 0, and our x is within δ of c--3200

it is within some interval of c - δ to c + δ--well, any interval...remember: rationals and irrationals are right on top of each other;3205

then that means our interval, this interval here, the one set up by 0 < |x - c| < δ, must contain both rational and irrational x-values.3224

If it contains both rational and irrational x-values, we know, because it has to be true for any x in this interval,3235

and since we contain both rational and irrational, we now have 1 and -1 popping out of our f(x).3246

Thus, the absolute value of 1 - l is less than 1/2, and the absolute value of -1 - l is less than 1/2 must be true simultaneously,3252

because we know that, within our δ bound, we are going to have both rationals and irrationals.3269

And anything within that δ bound, by the definition (for any ε greater than 0,3275

there exists δ greater than 0, such that, if you are contained within that boundary of δ,3279

it must map within that boundary of ε) that we are contained within that boundary of δ;3284

we are contained within c - δ to c + δ; but there are going to be rationals and irrationals in there.3289

So, that means that our ε boundary must contain both 1 and -1.3293

So, the absolute value of 1 - δ must be less than 1/2, and the absolute of -1 - δ must be less than 1/2.3299

But only one can be true; this and this are mutually exclusive, because you can't be in both bands at once,3304

if the band can only be a total distance of 1 wide around positive 1 or around -1.3312

1/2 down and 1/2 up don't manage to reach each other; they can't reach across, so you can't end up being in both bands at once with that ε of 1/2.3317

So, it says that they must be true simultaneously; but we know that only one can be true.3326

Since only one can be true, and we know that they have to be true simultaneously, those are impossible statements to be true at the same time.3330

And because we know everything else in our logic was flawless, except for our assumption that there existed some limit,3340

we know, therefore, that no δ can exist that will satisfy ε equal to 1/2; so no limit can exist.3347

And we have completed our proof.3365

All right, I think that is really cool; it is not for everyone, but I think it is so beautiful.3367

So, I hope you enjoyed it; I hope you get a cool sense of this.3372

And remember: what we talked about here is not something that is necessary for virtually any students--3374

certainly no student currently at a precalculus level.3381

If you are just in the level of this course, you don't need to know this stuff.3383

I am honest; I won't have to use this stuff at all.3386

It is going to maybe, maybe, maybe end up showing up when you take calculus, but probably not even then.3389

It is really something that you only need if you are getting into advanced math.3395

However, I think that this stuff is so cool; and there is no reason that, if you end up liking this stuff...3398

it is like going and checking out art for me; if you end up liking this stuff, you might as well check it out any time that you are able to enjoy it.3402

It is really cool stuff; it gets your brain thinking in all of these totally new ways.3409

And you can end up exploring this stuff for years and years and years.3413

You can make an entire career out of exploring this stuff.3416

If you think this is kind of cool, you can study mathematics in college, go on to become a mathematician, and just study math for the rest of your life.3418

There is a whole bunch of other stuff to explore; math is really, really cool.3425

All right, see you at Educator.com later--goodbye!3428