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Lecture Comments (10)

1 answer

Last reply by: Professor Selhorst-Jones
Tue Nov 19, 2013 10:55 AM

Post by Joel Fredin on November 13, 2013

Your teaching methods are insanely good. You are the god of teaching, seriously.

1 answer

Last reply by: Professor Selhorst-Jones
Fri Aug 16, 2013 4:12 PM

Post by Jorge Sardinas on August 15, 2013

for example 3 isn't 0/0 0 instead of 1 if not can you please say why Mr Jones?

3 answers

Last reply by: Professor Selhorst-Jones
Tue Aug 13, 2013 10:16 AM

Post by Taylor Wright on August 12, 2013

Please make a full series on Calculus! You are an amazing teacher!

1 answer

Last reply by: Professor Selhorst-Jones
Sun Apr 14, 2013 6:30 PM

Post by Orsolya Krispán on April 14, 2013

I can't stop loving your teachings methods! yeah, it really did freak me out at the 4th example. :D thank you guys for everything, you all really did help me a lot, and please cross your fingers for me to have a successful final exam and get into my dream university. Thanks a lot!

Idea of a Limit

  • This lesson marks our entry into an entirely new section of mathematics: calculus. From here on, this course will preview some of the topics you will be exposed to in a calculus class.
  • We can conceive of a limit as the vertical location a function is "headed towards" as it gets closer and closer to some horizontal location. Equivalently, a limit is what the function output is "going to be" as we approach some input.
  • For notation, we use

    lim
     
    f(x)
    to say, "the limit of f(x) as x approaches c."
  • Another way to conceive of a limit is to imagine "covering up" the location we're approaching with a thin strip. With that location covered, we ask, "Where does it look like this function is going (now that we can't see where it actually winds up)?"
  • Definition of a limit: If f(x) becomes arbitrarily close to some number L as x approaches some number c (but is not equal to c), then the limit of f(x) as x approaches c is L. Symbolically,

    lim
     
    f(x) = L.
  • In the above definition, there are two important things to note:
    • We are looking at f(x) as x→ c, but we are not concerned with f(x) at x=c.
    • When we consider x approaching c (x→ c), we are considering x approaching from all directions, not just one side. To have a limit, f(x) must go to the same value from both sides.
  • It should be pointed out that the above is not technically the formal definition of a limit. This will certainly be enough for now, but if you're curious to know about it, check out the next lesson, Formal Definition of a Limit. [But you won't need to understand that for a couple years (if ever), probably.]
  • Limits do not always exist. If the two sides don't settle down towards the same location, there will be no limit.
  • There are three main ways to find limits:
    • Graphs: Look at a graph of the function, and figure out if it makes sense to have a limit at the location. If so, find out what value the graph indicates. [This method is not very precise, but it gives a pretty good idea.]
    • Tables: Make a table of values where the input values get really close to the location we approach in the limit. If the output values settle down towards a single value the closer we get to the location, there is a limit. [This method is more precise and can give very good approximations, but is still not perfectly accurate.]
    • Precise Methods: Later on, we will see algebraic methods to find precise values for limits. The lesson Finding Limits will go over this in detail. For right now, though, we'll just stick to the methods of graphing and making tables.

Idea of a Limit

Fill in the table of values below (notice that you do not fill in the spot with the `?' symbol). Use your result to determine if the limit exists-if it does, estimate the limit.

lim
x→ 2 
4x−3

x
1.9
1.99
1.999
2
2.001
2.01
2.1
f(x)
             
             
             
    ?    
             
             
             
  • We fill in the table exactly like we would fill in any table of values. We have various x-values that we use as input, then we put down the corresponding output value. [Don't be confused by the fact that the table lists `f(x)' and we haven't explicitly named a function: the problem is implying that what comes after the `lim' is the function since it's the only thing that looks like a function.]
  • Plug in each value, see what the result is. Here are the first couple values:
    x=1.9     ⇒     4(1.9) −3     =     4.6

    x=1.99     ⇒     4(1.99) −3     =     4.96
    We do the rest of the x-values in the same manner, filling in the table as we go. Once completed, we have the below:
    x
    1.9
    1.99
    1.999
    2
    2.001
    2.01
    2.1
    f(x)
        4.6    
        4.96    
        4.996    
        ?    
        5.004    
        5.04    
        5.4    
  • Now that we have the table filled in, we can work on the limit. [If you didn't watch the video, make sure to do so. The idea of a limit is quite special and it's extremely important to understand the concept.] Remember, a limit is a way to ask what value the function approaches as x goes to a specific value. In this case, we have the limit as x→ 2, so we want to know what value the function is going to as x gets very close to 2. [Notice that we do not care what the actual value is at x=2. A limit is based on what happens near x=2, not at x=2. This is why there is a `?' symbol for x=2 on the table: because it does not affect the limit.] Look at the table to see what value the function is approaching. As x gets closer and closer to 2, it appears that f(x) → 5. The closer x is to 2, the closer f(x) is to 5. Therefore we have that the limit exists and that the limit equals 5.

x
1.9
1.99
1.999
2
2.001
2.01
2.1
f(x)
    4.6    
    4.96    
    4.996    
    ?    
    5.004    
    5.04    
    5.4    
Limit exists;    limx→ 2 4x−3   = 5
Fill in the table of values below (notice that you do not fill in the spot with the `?' symbol). Use your result to determine if the limit exists-if it does, estimate the limit.

lim
x→ −4 
x2+3x+3

x
−4.1
−4.01
−4.001
−4
−3.999
−3.99
−3.9
f(x)
             
             
             
    ?    
             
             
             
  • We fill in the table exactly like we would fill in any table of values. We have various x-values that we use as input, then we put down the corresponding output value. [Don't be confused by the fact that the table lists `f(x)' and we haven't explicitly named a function: the problem is implying that what comes after the `lim' is the function since it's the only thing that looks like a function.]
  • Plug in each value, see what the result is. Here are the first couple values:
    x=−4.1     ⇒     (−4.1)2+3(−4.1)+3     =     7.51

    x=−4.01     ⇒     (−4.01)2+3(−4.01)+3     =     7.0501
    We do the rest of the x-values in the same manner, filling in the table as we go. Once completed, we have the below:
    x
    −4.1
    −4.01
    −4.001
    −4
    −3.999
    −3.99
    −3.9
    f(x)
        7.51   
        7.05   
        7.005   
        ?    
        6.995   
        6.95   
        6.51   
  • Now that we have the table filled in, we can work on the limit. [If you didn't watch the video, make sure to do so. The idea of a limit is quite special and it's extremely important to understand the concept.] Remember, a limit is a way to ask what value the function approaches as x goes to a specific value. In this case, we have the limit as x→ −4, so we want to know what value the function is going to as x gets very close to −4. [Notice that we do not care what the actual value is at x=−4. A limit is based on what happens near x=−4, not at x=−4. This is why there is a `?' symbol for x=−4 on the table: because it does not affect the limit.] Look at the table to see what value the function is approaching. As x gets closer and closer to −4, it appears that f(x) → 7. The closer x is to −4, the closer f(x) is to 7. Therefore we have that the limit exists and that the limit equals 7.

x
−4.1
−4.01
−4.001
−4
−3.999
−3.99
−3.9
f(x)
    7.51   
    7.05   
    7.005   
    ?    
    6.995   
    6.95   
    6.51   
Limit exists;    limx→ −4 x2+3x+3   = 7
Fill in the table of values below (notice that you do not fill in the spot with the `?' symbol). Use your result to determine if the limit exists-if it does, estimate the limit.

lim
x→ 5 
x2−9x+20

x−5

x
4.9
4.99
4.999
5
5.001
5.01
5.1
f(x)
             
             
             
    ?    
             
             
             
  • We fill in the table exactly like we would fill in any table of values. We have various x-values that we use as input, then we put down the corresponding output value. [Don't be confused by the fact that the table lists `f(x)' and we haven't explicitly named a function: the problem is implying that what comes after the `lim' is the function since it's the only thing that looks like a function.]
  • Plug in each value, see what the result is. Here are the first couple values:
    x=4.9     ⇒     (4.9)2−9(4.9)+20

    (4.9)−5
        =     0.9

    x=4.99     ⇒     (4.99)2−9(4.99)+20

    (4.99)−5
        =     0.99
    We do the rest of the x-values in the same manner, filling in the table as we go. Once completed, we have the below:
    x
    4.9
    4.99
    4.999
    5
    5.001
    5.01
    5.1
    f(x)
        0.9    
        0.99    
        0.999    
        ?    
        1.001    
        1.01    
        1.1    
  • Now that we have the table filled in, we can work on the limit. [If you didn't watch the video, make sure to do so. The idea of a limit is quite special and it's extremely important to understand the concept.] Remember, a limit is a way to ask what value the function approaches as x goes to a specific value. In this case, we have the limit as x→ 5, so we want to know what value the function is going to as x gets very close to 5. [Notice that we do not care what the actual value is at x=5. A limit is based on what happens near x=5, not at x=5. This is why there is a `?' symbol for x=5 on the table: because it does not affect the limit. This is especially important in this case since f(5) does not exist because it would give [0/0].] Look at the table to see what value the function is approaching. As x gets closer and closer to 5, it appears that f(x) → 1. The closer x is to 5, the closer f(x) is to 1. Therefore we have that the limit exists and that the limit equals 1.

x
4.9
4.99
4.999
5
5.001
5.01
5.1
f(x)
    0.9    
    0.99    
    0.999    
    ?    
    1.001    
    1.01    
    1.1    
Limit exists;    limx→ 5 [(x2−9x+20)/(x−5)]   = 1
Fill in the table of values below (notice that you do not fill in the spot with the `?' symbol). Use your result to determine if the limit exists-if it does, estimate the limit.

lim
x→ 1 
2

x−1

x
0.9
0.99
0.999
1
1.001
1.01
1.1
f(x)
             
             
             
    ?    
             
             
             
  • We fill in the table exactly like we would fill in any table of values. We have various x-values that we use as input, then we put down the corresponding output value. [Don't be confused by the fact that the table lists `f(x)' and we haven't explicitly named a function: the problem is implying that what comes after the `lim' is the function since it's the only thing that looks like a function.]
  • Plug in each value, see what the result is. Here are the first couple values:
    x=0.9     ⇒     2

    x−(0.9)
        =     −20

    x=0.99     ⇒     2

    x−(0.99)
        =     −200
    We do the rest of the x-values in the same manner, filling in the table as we go. Once completed, we have the below:
    x
    0.9
    0.99
    0.999
    1
    1.001
    1.01
    1.1
    f(x)
        −20    
        −200    
        −2000    
        ?    
        2000    
        200    
        20    
  • Now that we have the table filled in, we can work on the limit. [If you didn't watch the video, make sure to do so. The idea of a limit is quite special and it's extremely important to understand the concept.] Remember, a limit is a way to ask what value the function approaches as x goes to a specific value. In this case, we have the limit as x→ 1, so we want to know what value the function is going to as x gets very close to 1. [Notice that we do not care what the actual value is at x=1. A limit is based on what happens near x=1, not at x=1. This is why there is a `?' symbol for x=1 on the table: because it does not affect the limit. This is especially important in this case since f(1) does not exist because it would give [2/0].] Look at the table to see what value the function is approaching. As x gets closer and closer to 1, we see that it shoots off to −∞ on the left side (x < 1) and shoots off to ∞ on the right side (x > 1). The value of the function is not converging to a single value as x→1: instead it is flying off in opposite directions. Therefore the limit does not exist.

x
0.9
0.99
0.999
1
1.001
1.01
1.1
f(x)
    −20    
    −200    
    −2000    
    ?    
    2000    
    200    
    20    
Limit does not exist: the function does not steadily approach a single value as x→1.
Using the associated graph below, determine if the limit exists. If so, give the value of the limit.

lim
x→ 1 
  −2x2+2x

x3−x2+x−1
  • The limit of a function is the value that the function approaches as the function goes toward a certain x-value. For this problem, we're interested in figuring out if the function is approaching a single value as x→ 1.
  • Look at the graph to see if the limit exists. Looking in the "neighborhood" around x=1, we see that both the left and right portions of the graph are being pulled towards a single value. Left of x=1 and right of x=1 are both heading towards where the empty circle is. Thus, because the function is clearly headed to the same place from both sides, the limit exists. [Notice that we do not care that the empty circle indicates the function does not exist at x=1 (it would give [0/0]). The limit only cares about what the function is headed towards, not what it actually is at the x-value. Thus we only care about what the left and right sides are headed towards.]
  • The value of the limit is what the function is headed towards as x→ 1. Previously, we saw that both sides are headed towards the empty circle. Thus, we only need to find what the height of the empty circle is to find the value of the limit. Carefully looking at the graph, we see that it is at a height of −1. Thus, as x→ 1, the function approaches a height of −1, so the limit has a value of −1.
Limit exists;    limx→ 1 [(−2x2+2x)/(x3−x2+x−1)]   = −1
Using the associated graph below, determine if the limit exists. If so, give the value of the limit.

lim
x→ 3 
2x−6

x2−4x+3
  • The limit of a function is the value that the function approaches as the function goes toward a certain x-value. For this problem, we're interested in figuring out if the function is approaching a single value as x→ 3.
  • Look at the graph to see if the limit exists. Looking in the "neighborhood" around x=3, we see that both the left and right portions of the graph are being pulled towards a single value. Left of x=3 and right of x=3 are both heading towards where the empty circle is. Thus, because the function is clearly headed to the same place from both sides, the limit exists. [Notice that we do not care that the empty circle indicates the function does not exist at x=3 (it would give [0/0]). The limit only cares about what the function is headed towards, not what it actually is at the x-value. Thus we only care about what the left and right sides are headed towards.]
  • The value of the limit is what the function is headed towards as x→ 3. Previously, we saw that both sides are headed towards the empty circle. Thus, we only need to find what the height of the empty circle is to find the value of the limit. Carefully looking at the graph, we see that it is at a height of 1. Thus, as x→ 3, the function approaches a height of 1, so the limit has a value of 1.
Limit exists;    limx→ 3 [(2x−6)/(x2−4x+3)]   = 1
Using the associated graph below, determine if the limit exists. If so, give the value of the limit.

lim
x→ 2 
 g(x),    where g(x) =



2x,
x ≤ 2
1

2
x + 5,
x > 2
  • The limit of a function is the value that the function approaches as the function goes toward a certain x-value. For this problem, we're interested in figuring out if the function is approaching a single value as x→ 2.
  • Look at the graph to see if the limit exists. Looking in the "neighborhood" around x=2, we see that both the left and right portions of the graph are being pulled towards a single value. Left of x=2 and right of x=2 are both heading towards the top "corner" of the graph. Thus, because the function is clearly headed to the same place from both sides, the limit exists.
  • The value of the limit is what the function is headed towards as x→ 2. Previously, we saw that both sides are headed towards the "corner" at the top. Thus, we only need to find what the height of that location is to find the value of the limit. Carefully looking at the graph, we see that it is at a height of 4. Thus, as x→ 2, the function approaches a height of 4, so the limit has a value of 4.
Limit exists;    limx→ 2  g(x)   = 4
Using the associated graph below, determine if the limit exists. If so, give the value of the limit.

lim
x→ −3 
  (x+1)3

30(x+3)
  • The limit of a function is the value that the function approaches as the function goes toward a certain x-value. For this problem, we're interested in figuring out if the function is approaching a single value as x→ −3.
  • Look at the graph to see if the limit exists. Looking in the "neighborhood" around x=−3, we see that the left and right portions of the graph are going to totally different places. For the part of the graph that is left of x=−3, it's flying off towards ∞, while the part that is right of x=−3 is flying off towards −∞. Thus, because the two sides of the function do not "agree" on a single value they are both headed towards, the limit does not exist
  • Because the limit does not exist, it has no value. We cannot assign a value to something that does not exist.
Limit does not exist.
Draw a graph of the below function, then use that to determine if the limit exists and, if so, what its value is.

lim
x → 3 
 f(x),    where f(x) =



−2x+3,
x ≤ 3
(x−2)2 − 4,
x > 3
  • Begin by drawing a graph of the piecewise function g(x). [If you're unfamiliar with piecewise functions or need a refresher, check out the lesson Piecewise Functions in the section on Functions. That lesson will fully explain the idea of a piecewise function along with showing how to graph them.]
  • When graphing the piecewise function, you can create a table of values to help you see how to plot points and create the graph.

    Another alternative is to create another graph that plots each "piece" of the function without regard to the interval the "piece" is graphed on. This makes it easier to see how each "piece" is a part of the whole graph of the piecewise function. We can see this below, where −2x+3 is graphed in red and (x−2)2−4 is graphed in blue. Once you see how each "piece" is graphed, erase everything except the interval that it is used on. For example, the right side of −2x+3 will be erased so the only part remaining is for the section where x ≤ 3. We can see the completed graph of the piecewise function f(x) on the next step.
  • However you create the graph of f(x), you should get the below [The graph below has been color-coded to help you see each individual "piece" that makes it up. This is not necessary, since the graph of f(x) is both "pieces" taken together, but it does make it slightly easier to understand how f(x) creates its graph.]:
  • Now that you have a graph of f(x), we want to consider if it has a limit as x→ 3. To decide this, we need to see if the function is approaching a single value from both sides in the neighborhood around x=3. Looking at the graph, we see that both the part to the left of x=3 and to the right of x=3 are headed to the same location. Therefore the function has a limit as x→ 3. The height of the location being approached is −3, so that is the value of the limit.
Limit exists;    limx → 3  f(x)    =  −3
Below is the graph of g(x) = {
x2 −8,
x ≤ −1
x2 − 8,
x ≥ 1
. Explain why limx→ 0 g(x)  does not exist.
  • Start off by noticing that g(x) has a gap in its graph. There's nothing on the graph for the x-interval (−1, 1). Why? Because the function is not defined over this interval. Look at how the piecewise function is defined. We know what to do for x ≤ −1 and we know what to do for x ≥ 1, but it says nothing for −1 < x < 1. Thus, on that interval, the function is not defined so it does not exist for any of the values in that interval.
  • Next, consider that a limit is the value that a function approaches as the x-value approaches some specific number. For this problem, we're interested in seeing what value g(x) approaches as x→ 0
  • Finally, notice that the function is not approaching any value as x→0. While we can "see" what happens for x ≤ −1 and x ≥ 1, because the function is not defined for (−1, 1), there's simply nothing happening inside the gap. Because of this, there is no limit point to approach. The function is not defined in that "neighborhood", so there is no limit. [Notice that this is different from the function not being defined at only one point. While a limit is unaffected by what the function's value (or lack of value) is at the specific x-value being approached, it still must be defined in the vicinity around that x-value.]
The function is not defined on the x-interval of (−1, 1), so it is impossible for the function to approach any values while using that interval, and thus no limits can exist for x-values in that interval.

*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.

Answer

Idea 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 0:05
  • Motivating Example 1:26
  • Fuzzy Notion of a Limit 3:38
    • Limit is the Vertical Location a Function is Headed Towards
    • Limit is What the Function Output is Going to Be
    • Limit Notation
  • Exploring Limits - 'Ordinary' Function 5:26
    • Test Out
    • Graphing, We See The Answer Is What We Would Expect
  • Exploring Limits - Piecewise Function 6:45
    • If We Modify the Function a Bit
  • Exploring Limits - A Visual Conception 10:08
  • Definition of a Limit 12:07
    • If f(x) Becomes Arbitrarily Close to Some Number L as x Approaches Some Number c, Then the Limit of f(x) As a Approaches c is L.
    • We Are Not Concerned with f(x) at x=c
    • We Are Considering x Approaching From All Directions, Not Just One Side
  • Limits Do Not Always Exist 15:47
  • Finding Limits 19:49
    • Graphs
    • Tables
    • Precise Methods
  • Example 1 26:06
  • Example 2 27:39
  • Example 3 30:51
  • Example 4 33:11
  • Example 5 37:07

Transcription: Idea of a Limit

Hi--welcome back to Educator.com.0000

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

This lesson marks our entry into an entirely new section of mathematics; we are entering calculus territory.0005

From here on, this course will preview some of the topics that you will be exposed to in a calculus class.0011

You might wonder what calculus is and why it matters.0016

In short, calculus is a new way to look at functions.0021

It gives us new tools to analyze function behavior and see how they relate to one another and the world around us.0025

As to why it matters: calculus is crucially important to science and engineering.0031

If you want to get anything of any real depth done in those fields, you need calculus in your tool belt.0036

Plus, I think it is just really cool; it is one of these really cool things that you can do in mathematics.0041

It is a really new, interesting idea, and we get to play around with it and do all sorts of cool stuff.0045

And it creates...from here one, we get to see a whole bunch of new stuff that lets us explain a whole bunch of phenomena in the real world--really cool stuff.0050

For the next few lessons, we will focus on limits.0057

Limits allow us to describe functions in ways that were previously impossible for us to describe.0060

We will be able to approach the infinite and the infinitesimal (that being the infinitely small) things with them.0065

And they make up the heart of calculus; they are pretty much the foundation that the rest of calculus rests upon.0071

So, it is really useful to have a good understanding of what a limit is and how it works,0077

before we can start moving into other ideas in calculus that have more direct applicability to the daily world.0080

All right, let's start with a motivating example to get things started.0087

Consider the function f(x) = x/x: what happens if we try to look at f(0)?0091

Well, if we plug 0 in, then we are going to get f(0) = 0/0; but since dividing by 0 is not defined, f(0) does not exist.0098

Therefore, we cannot evaluate the value of the function at 0; f(0) is not a thing.0110

We can't do anything with it, and that is the end of the story--that is it.0114

And up until now, that would have been the end of the story.0119

But now, when we look at f(0), well, OK, it is kind of the end of the story on what f(0) is.0122

But before we entirely refuse the idea of considering f at 0, considering how f and 0 interact, let's look at the function's graph.0131

If we look at f(x) = x/x, look at that: the function is always 1.0139

Now, right here at x = 0 on our vertical y-axis, we have this hole here.0145

That circular hole tells us that it does not actually exist there.0152

But anywhere that isn't x = 0, we end up getting a 1 out of it.0155

On the one hand, f(0) does not exist; you plug in a 0; you get 0/0; that is bad.0160

You are not allowed to divide by 0, so we say that f(0) does not exist.0166

But on the other hand, it is obvious where it is headed: look, the thing is going right in towards that 1.0169

On the one hand, sure, it doesn't exist; but it totally should be a 1.0176

So, we cannot say f(0) = 1, because f(0) does not exist, remember.0182

But we still want some way to talk about where it was headed.0188

It was clearly going to be a 1 before those pesky rules about dividing by 0 got in the way and stopped us0192

from being able to figure out what the real answer should have been--0199

what we feel like it was going towards, at least--perhaps not the real answer, because...0203

we will talk about it; you will see other things about why we can't really assign a direct value to it.0208

But we can talk about this idea of where it was headed; and we talk about that with a limit.0212

With this idea in mind, we want to think of a limit as where it was headed.0219

A limit is the vertical location that a function is headed towards--what it is going to be at as it gets closer and closer to some horizontal location.0223

In that previous example, our motivating one of x/x, we had that, as it got closer and closer to 0, it got closer and closer to being at a height of 1.0236

In fact, it was always at a height of 1, no matter where it was.0244

We end up seeing this sense of...as it gets closer to 0, it is really, really close to this value, where it is headed towards.0248

Equivalently, a limit is what the function output is going to be as we approach some input.0255

As we get close to some input, what output seems like it is going to come out of it--what should it be; what is it going to be?0263

We are going to be talking about this idea a lot, so we want some notation to describe it.0273

We use this notation right here: this part here, the "lim," says that it is the limit we are working with.0278

The x arrow c says what we are going towards; and the f(x) is the thing that actually is having the limit be applied to it; that is the function.0285

This says "the limit of f(x) as x approaches c"; so as x gets close to c, what happens to f(x).0292

You might also hear this spoken aloud as "the limit as x goes to c of f(x)"; I tend to say that a lot.0302

Or you might hear some similar variant; in any case, the idea is the same.0308

It is this question of what f(x) does as x gets very close to c.0311

x is going to c, and our question is, "What will f(x) do in response to x going to c?"0317

Let's test out our new idea on an old friend, f(x) = x2, the good old parabola.0326

Consider if we wanted to find what value x2 approaches as x approaches 2.0332

As x gets closer and closer to 2, what does x2 become?0337

Well, if we graph f(x) = x2, we see that the answer is exactly what we would expect.0342

2: as we get really close to a 2, we end up getting really close to a 4.0349

As we come in from the left side, we can see that the value we are getting to, as we get closer and closer0356

to a 2 from the left side--our vertical height gets closer and closer to a 4.0363

Similarly, if we come in from the right side, as we get closer and closer to 2, and x = 2, from the right side,0370

on our horizontal location, we get closer and closer to a vertical location of 4.0377

We end up getting 4 as the limit: as x approaches 2, x2 gets really close to 4.0382

It becomes 4 as x approaches infinitely close to 2.0388

So, as x gets closer and closer to a value of 2, f(x) gets closer and closer to a value of 4.0393

And that is why we end up getting this limit.0402

Now, to expand our ideas, let's modify the function and see what happens.0405

Consider the piecewise function g(x), which equals x2 when x is not equal to 20409

(that is most of our parabola right here) and 1 when x equals 2.0418

At the specific value of x = 2, as opposed to following our normal parabolic arc, we end up putting out just a value of 1.0424

This is a piecewise function; if this is not sounding familiar, and you have no idea how to interpret this sort of thing,0433

go back to the lesson...we saw this a long time ago, near the beginning of this course...the lesson on Piecewise Functions.0439

They will show up a lot, especially in the beginning of calculus.0444

They are an important thing to have in your understanding.0447

So, if this doesn't make any sense, go and check out piecewise functions from the early part of the course, when we were studying functions.0450

All right, with this idea in mind, let's ask what the limit is, as x approaches 2, of g(x).0456

Well, what we want to know--we might be tempted, first, to say, "Look, it is right here; that is what x is at 2."0463

Yes, that is what g(x) is at 2; g(2) is 1.0469

Well, yes, but we can also think about it as what we are getting close to.0476

Well, as we get close to x = 2 from the left side, we see that we are getting really, really close to what height?--the height of 4.0482

As we approach from the right side, we see that we are getting really, really close to what height?--the height of 4.0492

So, our first automatic response might be to think that it has to be here, because that is what it is at x = 2.0499

But the limit isn't about where you actually are; as a limit, x/x, as x goes to 0, produces 1,0505

even though actually the function just fails to produce anything.0516

A limit is about where you are headed towards, not what actually happens at that location.0519

We don't really care about what happens here; this part isn't important.0525

The thing that actually happens at 2 isn't important; it is a question of what happens on our way to 2.0529

Well, on our way to 2, the thing that we are getting close to is this location at 4.0535

That is the location we are getting to; so the answer for this limit will be 4.0541

So, g(2) is not equal to 4; g(2) is equal to 1, because at x = 2, we just put out 1 for that function.0547

But as x approaches 2, the value of g(x) approaches 4.0555

g(x) jumps; it does this sudden leap off of the parabolic arc, only at x = 2; it only swaps at x = 2.0561

What it seems to do, up until that moment, is behave like x2, because for everything that isn't x = 2,0571

for everything other than x = 2, it behaves just like x2.0579

Up until that moment, it is behaving just like x2; and because of this, it has the above limit of becoming a 4.0583

It is going towards a 4 until that single moment where all of a sudden, when it actually touches the 2, it jumps away.0590

But up until that moment, it seems like it is going there; so that ends up being our location.0596

The vertical height that it seems to be going to as x approaches 2 is 4.0601

In a way, we can visualize what is going on by doing the following:0608

Begin by graphing the functions normally; we graph f(x) normally, x/x...oops, that should not say x/x;0611

that should be x2; I'm sorry about that; that should be f(x) = x2.0618

And over here, g(x) = x2 when x is not equal to 2, 1 when x equals 2.0624

We have this single part that jumps away--this single point that is not on the normal curve.0630

Then, what we can do is end up covering up the part that we are going to.0634

Notice: in this case, what we are about to consider is the limit, as x approaches 2,0640

as we get close to this value, as we get really close to this on both of them.0645

With that idea in mind (I'm going to swap this back out; it should be an x2; we are looking at the limit as x goes to 2),0651

as we take the limit, what we do is cover up the horizontal location that x is approaching,0661

because, since it is a limit, what we are concerned with is what happens on our way to that value.0668

But we don't actually care about that value in specific.0673

We don't care about that horizontal location; what we care about is our way to the horizontal location.0676

It is the journey that matters, not the destination, when it comes to limits.0682

So, the limit is the height we expect--what we feel like would happen without peeking under the cover.0686

We have that black bar there that keeps us from being able to see what it actually turns out to be.0694

But in this case, it seems like what it is going to come out to be is 4.0699

If all of the information we have is just the picture in front of us (except that part that has been covered up--0703

we are not allowed to look under it), the information that we have makes it seem like it looks like it is going to 4 in both of them.0707

The idea is the question of what we expect will end up happening.0713

Where does it seem like this function is going to? That is what a limit is about.0717

With all of that in mind, we now have the ability to create a more formalized definition.0722

The definition of a limit: if f(x) becomes arbitrarily close to some number l, as x approaches some number c,0727

but is not equal to c (we don't actually care about that horizontal location;0735

we just care about the way to that horizontal location), then the limit of f(x), as x approaches c, is l.0739

Symbolically, we write that as "limit as x goes to c of f(x) equals l"; the limit of f(x) as x approaches c is l.0747

So, as x gets close to c, what value does our f(x) seem to go to--what value is f(x) getting towards as we get close to that horizontal location?0757

There are two important things to note: we are looking at f(x) as x goes to c, but are not concerned with x = c.0769

So, for the purposes of a limit, x is never equal to c; we are only concerned with what happens on the way to c, but not actually x at c.0777

Remember: it is the journey, not the destination, that matters when it is a limit.0786

The other thing to notice is that, when we consider x approaching c, we are considering x approaching from all directions--not just one side.0790

It is not just this side; it is not just this side; it is both sides coming together.0798

To have a limit, f(x) must go to the same value from both sides.0804

The right side and the left side have to agree with each other.0809

If they go to totally different things for a horizontal location, if they go to totally different heights, then they don't agree; there is not a limit.0812

There is not a sense of what to expect if they are going to totally different places.0819

Where is it going to be? Is it going to be somewhere in the middle?0824

Is it going to be the top one? Is it going to be the bottom one?0825

We don't have a good sense of which side to trust, so we can't get a limit out of it.0827

The two sides have to agree for us to end up having a limit.0832

Technically, I want to point out that this isn't actually the formal definition of a limit.0836

That said, what we are seeing here is going to certainly be enough for now.0841

This idea, the definition of a limit that we just talked about (this whole thing here and all of our commentary that came after it--0846

all of the stuff that we have been working through so far in this lesson)--this is plenty for the class you are currently in.0852

A precalculus-level course like we are in right now--this is more than enough of an understanding of what a limit is.0857

You are doing great at this point.0862

Even for a calculus-level class, this is really pretty much enough understanding.0864

Some courses will maybe vaguely talk about the formal definition,0868

but very, very few will really expect you to fully understand the formal definition of a limit.0872

This is really all that they are looking for--this sense of a limit being what you are going towards, but not where you actually end up.0877

And for pretty much most science courses, most engineering courses, this is really all you need.0885

If you want to talk about the really formal definition of a limit, that is going to show up later on,0889

in really advanced math courses, like second--maybe even third-year college math courses, really proof-heavy math stuff.0894

And I think that that is really great stuff.0901

But for the most part, you will be fine with just this, probably forever.0903

However, if you are really interested in mathematics, if you are curious, you might want to check out the next lesson, Formal Definition of a Limit.0907

I think that this stuff is really, really cool; and I have a great lesson that will help us explain and understand what the formal definition of a limit is.0913

But frankly, probably 99 times out of 100, you are never going to need to know that stuff.0920

Pretty much any class that you will be taking in the next two years is never going to actually require you to know the formal definition of a limit.0924

So, don't worry too much if you don't feel like watching it; it is totally a fine lesson to end up skipping.0930

But if you are interested, it is really cool; and if you are interested in this, you might end up being interested in taking advanced math classes later.0935

And it will totally come into play later on, and you will be a step ahead of everybody else in understanding this fairly complicated idea.0940

All right, let's keep talking about limits: so far, all of the examples we have seen have had limits.0946

But a limit does not always exist: for example, consider the limit as x goes to 0 of 1/x.0952

Well, here what we are doing is looking at as x goes to 0 (the y-axis is at x = 0).0958

As we come in from the right side, we aren't actually going to a single value.0965

We are just going to go up and up and up and up and up and up and up and up, and we don't ever stop going up.0969

It is a vertical asymptotes; when we worked on rational functions and vertical asymptotes, we just went on up forever.0975

There is no single l value, no single limit number that we are going towards.0980

So, there is nothing that can be agreed upon.0985

And even worse than that, we end up going in the totally opposite direction when we come from the left side.0987

One side is shooting off to positive infinity; the other side is shooting down to negative infinity.0991

There is nothing to say for the limit here.0996

They are not approaching something where they are agreeing on some value.0997

They are not even going to separate values; they are just blasting off to infinity on both sides.1001

We don't really have a good way to talk about this; there is no way to assign a specific number value that we expect will happen at x = 0,1005

because it is just going to clearly go crazy, and that is that; there is nothing that it is going to go to.1012

So, because of that, we say that the limit does not exist.1016

This limit does not exist; there can be no limit, as x goes to 0, because there is no single value that 1/x is headed towards.1019

There is nothing that we are going to end up seeing them agree on.1028

Even one side is not going to agree on anything, because it just keeps going up.1032

There is no single value; therefore, there is no limit that we get out of it.1035

Still, I want to point out: for x approaching any other value than 0, the limit would exist, because it would approach a single value.1039

If we approached 1 as our thing, then we would end up approaching 1.1048

If we approached 4 as our thing, then we would end up approaching 1/4.1053

If we went to -3, if we were approaching -3 from both sides, then we would be going to -1/3.1059

All of those make sense; the only issue is here at x = 0, where, because it has an asymptotic thing, it just goes crazy and shoots off in both directions.1066

There is nothing that we can end up pinning it down with, so we have to say that the limit does not exist.1073

But anywhere else on 1/x would exist; but x going to 0 does not exist for 1/x.1077

In the previous example, the limit as x goes to 0 of 1/x, that didn't exist.1084

But at the same time, 1 divided by 0 does not exist, either; f(0) didn't exist, and so limit as x goes to 0 of f(x) didn't exist.1089

And there is a connection there; however, it is possible for the function to exist while the limit does not exist.1097

So, to see this, take, for example, g(x) equals the piecewise function x2 when x is less than 1.1105

(we see this parabola on the left side) and 4 - x when x is greater than or equal to 1 (and we see this straight line on the right side).1112

We have every single point in the real numbers defined.1121

They will end up having some value that the function will put out.1124

If you plug in any number, it is going to either be on the straight line portion, or it is going to be on the parabola portion.1128

So, g(1) = 3; if we plug in 1, we use 4 - x; 4 - 1 gets us 3; that is this point right here.1134

However, the limit as x goes to 1 of g(x) does not exist; why?1148

Well, from the left side, we end up approaching this value here.1153

We are approaching this side as we approach from the left side.1157

However, as we approach from the right side, we end up approaching this totally different value.1160

And so, there is this big gulf between the two limits.1166

We are going to two totally different places at this horizontal location.1169

So, the function approaches two totally different values from the right and left sides.1173

Since they don't agree, we can't say, "Oh, that is what we expect," because we have totally different expectations from the two sides.1180

So, that means that the limit does not exist.1186

How do we actually go about finding limits in general?1190

The first really great way to do this is with graphs.1192

One way to find the value of a limit is just to look at a graph of the function.1195

If you have a graphing calculator, you can plot it on the graphing calculator.1198

If you have some sort of graphing program, like I make these graphs with, you can plot it on a graphing program.1201

Or if you are just really good at making graphs, you can draw a graph.1205

Figure out if a limit there makes sense: is there something where both sides are going to the same thing?1208

And if so, find what value the graph indicates.1214

For this case, we have f(x) = (x2 + x - 2)/(x2 - x);1217

(x2 + x - 2) on the top, (x2 - x) on the bottom of the fraction.1223

So, we see that there are some parts where we have issues.1227

If we plug in x = 0, we end up having an asymptote; and if we plug in x = 1, we end up having this hole here.1230

But we can still ask what the limit is as x goes to 1.1237

Well, if we go to the graph, as x goes to 1 from the left side, we end up seeing that we are approaching that height of 3.1241

As x goes to 1 from the right side, we see that we are approaching the exact same height.1249

We end up getting up a value of 3; either way we approach this, we are going to end up seeing that 3 is what the expected value is.1255

The graph shows us that we are working towards 3, or something at least on this graph looks very close to 3.1263

So, we could say 3; but of course, we are reading a graph.1267

Reading a graph isn't always as precise as we would like.1271

Reading a graph, we know that you can sometimes be off by 1/2 or 1 whole thing.1275

So, it doesn't give us a perfect answer; but it gives a pretty good idea.1279

We have a good sense of what it is going to be, although it is not perfectly precise.1283

However, limits do have the massive benefit of being able to allow us to get an intuitive sense of how the thing is working.1287

They will let us see what the function is, as a general idea; and sometimes that is the most useful thing of all.1293

Graphs are really, really handy, even if they don't let us see precisely what the value is;1297

they let us understand what is going on--does it even make sense for it to have a limit here?1301

Things like that are what a graph allows us to answer.1305

Alternatively, if we want something that is more precise, if we want a more precise sense of where the limit will go,1309

or we don't want to graph the function, just because we have some sense of what the graph looks like,1315

or we just don't feel like graphing it, because we know it is going to be a pain,1318

but we have enough of an idea to know that the limit would exist there; we can use a table of values,1320

where x will approach the value that it approaches in the limit.1326

Once again, we have the same f(x) = x2 + x - 2 over x2 - x.1331

And now, we are looking at the limit as x goes to 1 of that function, the same limit as before.1335

What we can do is: we have 0.9, 0.99, 0.999...we are approaching 1 from the left side there.1341

We can approach it from the right side: 1.1, 1.01, 1.001...we are getting closer and closer values.1350

Now, of course, we can't actually plug in x = 1, because if we plug in x = 1, it is just going to fail on us.1357

The limit isn't about where it actually would be; it is about what happens on the way there.1364

So, we don't plug x = 1 into our table; all we are concerned about is what happens to the numbers as they get really close to x = 1.1368

We calculate this with a calculator: .9 comes out to be 3.222; .99 comes out to be 3.020; .999 comes out to be .002.1375

Going from the other side, 1.1 is 2.818; 1.01 is 2.980; 1.001 is 2.998.1385

So, we can see that the value that we are getting close to, 2.998, 3.002...we are clearly tending pretty close to the value of 3.1393

So, we can assign that this limit is going to end up having a 3 as we get closer and closer and closer.1403

Now, maybe we are off by .0001 or some small number; but we can be pretty sure that,1409

unless it does some really sudden jump there, 3 is probably going to be pretty close to it.1416

That lets us get a good approximation--probably a good approximation within many decimal places, but we are not absolutely, precisely sure.1421

Still, that is really, really close; and the closer our table has x approach the limit, the more sure we can be.1429

If we, instead of using 1.001, would use 1.0000000001, we would be that many more decimal places sure of where we are headed.1438

The same happens with 0.999999999; by plugging in more and more decimal places, we get more and more accuracy.1448

So, we can be more and more sure of what the value that we will end up getting out of that limit is.1456

If you have a graphing calculator, this is a great use for the table of values feature,1460

where you can just set up a function, then go the table of values,1466

and have it be an independent thing where you plug in each number.1469

You plug in .9, .99, .99999; you hit them all in, and it will just put out values for you.1471

And you won't have to type in the entire expression over and over.1476

So, check out the course appendix; there is that appendix on graphing calculators as part of this.1480

It is a great thing to check out if you have access to a graphing calculator.1485

That table of values will make your life so much easier when you are working through this sort of thing--this is a really good use here.1487

Precise methods: the lesson after the next is finding limits (not the very next lesson,1494

which will be the formal definition of limits, which you can totally skip, if you are not particularly interested;1499

but the one after that--you will want to watch it, and that is Finding Limits).1504

In it, we will see ways to precisely find the limit of a function through algebraic methods.1508

Graphing and a table of values--those two things give us really good approximations.1512

They give us a good sense of what is going on, but they don't tell us what the value has to be, precisely.1517

We will figure out algebraic methods in the next, next lesson, Finding Limits.1521

For right now, though, we will just stick to the methods of graphing and making tables: they are pretty good methods, actually.1526

Even after you learn those more precise methods about how to figure out algebraically, precisely, don't forget about these.1532

They can be really handy when you can't figure out how to figure it out precisely, algebraically,1538

but you still need to evaluate it and get a good sense of where it is going.1543

You can always use these methods.1546

Now, often you will end up having problems where the problem says that you have to get it precisely, and you won't be able to use these methods.1549

But sometimes, you will have some really, really complicated thing, and you won't be able to work it out.1555

You can just put it in a table of values, something like this, and you will be able to get a good sense of where it is headed; and that can be useful.1559

All right, we are ready for some examples.1566

For each limit below, if it exists, determine the value using the associated graph.1568

Our first one is the limit as x goes to 0 of 1/x2.1571

Well, what we can see is that, as we get closer and closer to x = 0, this thing just shoots off; both sides fly off to infinity.1576

They shoot off forever and ever and ever.1585

So, even though they are both going towards positive infinity, do they ever end up establishing at a single value?1587

Does it ever settle on a single value? No, they are always going to keep going up.1593

It is going up asymptotically, so there is never a number that they get steady out on.1598

They never decide to land on 47 or some other number.1603

Since they keep going on forever, it doesn't have a limit.1606

The limit here does not exist, because it never settles on a single value.1609

It shoots up to infinity forever, and since it never ends up staying around at a single value, it never settles on a single value.1619

So, we end up not being able to get a limit out of it.1628

The next one is the limit as x goes to -3 of (x + 3)/(x2 + 5x + 6).1632

In this one, if we plug in -3, look: there is a hole, so we can't just directly plug in -3.1639

But from the graph, we can see that, yes, there is nothing wrong.1645

We are approaching the same thing from the left and the right side.1648

What value are we approaching? We are approaching a value of -1, so the limit is -1; great.1651

The next one: Evaluate the limit as x goes to -2 of 3x2 + 5x + 1.1659

We could graph this; this is just a parabola--it wouldn't be too tough for us to graph.1664

And if we graphed it, we would end up seeing that it looks something like this.1669

And we could figure out what it is--draw a really careful graph; but drawing a really, really careful graph takes a fair bit of effort.1676

And we have a good sense that there are going to definitely be limits everywhere,1682

because it never does anything weird; it never jumps around; everything that we expect to happen is what happens.1685

So, we don't have to worry about that; but that is what the graph gets us.1690

At this point, now we want to actually figure out what the value is.1693

The easiest way to figure out what the value is: we can just plug in values; we make a table of values.1695

We are going to have x on the left side, and the f(x) that comes out on the right side; what value are we approaching?1702

We are approaching -2; so x going to -2...if we are a little below -2, we will be at -2.1, then -2.01, then -2.001.1708

On the other side, we will be coming away from -2: -1.999, -1.99, -1.9...1722

If we plug these things into our calculator, -2.1 gets us 3.73; -2.01 gets us 3.0703; -2.001 gets us 3.007003.1732

Flipping to the other side, -1.9 would get us 2.3; -1.99 would get us 2.9303; -1.999 gets us 2.993003.1748

So, it is pretty clear what we are approaching.1763

As we go in from each side, -2.001, -1.999...they are getting really, really close to this middle value of 3.1765

So, that is what the limit ends up being; the limit comes out to be equal to 3; that is the value that it is approaching.1774

Also, remember how we talked about...well, look at that graph: the way that that graph is there--it doesn't do anything weird.1781

Everything that we expect to come out of it is what is going to be that function's location.1789

So, another thing that we could do is: because, in this specific case, the graph doesn't do anything weird,1794

the function doesn't jump around in any weird ways; there are no breaks; and there is nothing strange about it.1799

So, if there is nothing strange about it, what we could do is: we could also say,1806

"Well, that means that the limit has to be what the function actually ends up going to at that point."1808

So, we could also just evaluate it by plugging it in: 3(-2)2, plus 5 times -2, plus 1.1813

3 times...-2 squared gets us 4, plus 5 times -2 gets us -10, plus 1...3 times 4 is 12, plus -10...so 2 + 1...1821

That came out to be 3 as well, so that checks out, as well.1830

So, what we just saw there was actually one of the precise methods of doing this stuff.1834

We will talk about this more in Finding Limits; but it seemed like a really easy one for us to see...1838

we are going to start to get a sense of how this stuff works, and we want to find the precise stuff.1842

So, we will see more, and we will also understand exactly why we can do this, in the coming lessons.1846

All right, the third example is the limit as x goes to 0 of sin(x)/x.1851

If we were to graph this, if we used a graphing calculator to graph what comes out of this, we would see that it is going to look something like this.1856

Oops, I actually made a mistake with that graph; we would see that it was going to look something like...there we go; that is better.1865

It is going to look something like that.1877

The easiest way to do this is to just do a table of values; we have x, and f(x) is coming out of it, as we plug in various values for x.1880

So, our x is approaching 0; if we are approaching 0 from under it, we are going to be at -0.1, then -0.01, then -0.001.1890

And you could use different numbers, as long as they continue to get closer and closer to 0.1902

But I think those are pretty easy ones to use.1905

From the other side, we would be coming away: 0.01 on the positive side now; 0.01, and 0.1, now that we are past the 0.1908

We plug these into a calculator; we figure out what it is; we get...-0.1 going in gets us 0 point...1917

Also, remember: this has to be in radians for us to...1922

Actually, it doesn't have to be in radians because of this specific problem.1925

But any time we end up seeing a function, we should assume that it is in radians,1927

unless we have been explicitly told otherwise--that it is in degrees.1931

But normally, assume that it is in radians when you are doing math.1934

0.998334...plug in -0.01: 0.999983; -0.001: 0.9999999.1938

On the other side, there is 0.1: we have 0.998334; for 0.001, it is 0.999983; and for 0.001, 0.999999.1957

It is pretty clear what we are ending up approaching here.1976

As we get closer and closer, the value that we are approaching is 1.1978

The value that this is getting really close to is 1, and it seems to show that it is going to get really close to it.1981

So, we see that the value of this limit is 3.1988

The fourth example: Using the associated graph, explain why the limit below does not exist: the limit as x goes to 0 of sin(1/x).1992

Just glancing at this thing, the answer is simply that it goes crazy.2002

Look at this thing that is happening here--it is going crazy!2009

This graph is doing some really weird stuff as x gets close to 0.2014

So, as x gets close to 0, what is going on?2019

Well, notice how we can see that it is going up, and then it goes all the way down.2021

And then, it goes all the way up, and then it goes all the way down.2026

And then it goes all the way up and all the way down; the same thing happens on the other side:2029

up and then down, and up and then down, and then up and then down.2032

Well, what it seems to be doing is going up/down, up/down, up/down, faster and faster, as it gets closer and closer.2035

It is going crazy; it is bouncing up and down forever as it gets closer to this x going to 0.2040

It is just bouncing so, so fast; so as we get closer and closer to that x going to 0, we don't have any idea what to say,2053

because we are flying around; the closer you get to x = 0...it constantly changes.2060

The function is constantly changing, constantly going up and down.2065

Because it is constantly bouncing up and down forever and ever and ever and ever, we end up saying that it doesn't have a limit.2068

It doesn't exist; the limit does not exist, because it doesn't settle on anything.2074

A limit has to be settling towards some value, and it has this craziness there.2078

So, we end up not being able to say it has a limit; the limit does not exist.2083

If you want a better idea of what is going on here (and I think it is always cool to have a better idea of what is going on),2088

we can break this into two pieces: the graph of 1/x ends up graphing (let me put that in black,2092

just so we can easily see it)...here is our axis...it ends up looking like this.2099

We are used to that nice vertical asymptote there.2110

And then, the graph of sin...let's say t, just some dummy variable t that we plug in...it is going to end up having that nice periodic graph.2112

That is how sine works: it goes up and down and up and down and up and down and up and down.2124

That is what happens; however, what we have here, the t that is going into this, is 1/x.2130

So, what happens is that the value, as we get closer and closer to 0...2138

well, what happens to that asymptote as we get closer and closer to 0?2144

As we get closer and closer to 0, they shoot off to infinity.2148

So, what we have is: we effectively have an infinitely long amount of stuff that we are plugging into sin(t) in a very small, compact space.2151

That is why we end up seeing that it is going slowly.2160

And then, as it gets closer and closer to 0, 1/0 (that is not really formally accurate, but)...1/0 is effectively shooting off to infinity.2163

1/0 shoots off to infinity; so if we are shooting off to infinity, and then we are plugging something2173

that is going all the way out to infinity into sine, well, that means our periodic thing is going to go up and down forever.2178

But because we are doing all of this forever-ness in all of this very tiny distance of .1 to 0,2184

that means we end up having it go faster and faster and faster and faster, because it has to manage2190

to do all of infinity in 0 to 0.1--pretty crazy.2194

That is why we end up seeing this behavior.2200

And if that still doesn't quite make sense, don't worry.2202

Honestly, you will be fine in a precalculus class, and even a calculus class, without fully understanding this idea.2204

But if you let this rattle around in your head--think about it for a while--you might start to think, "Oh, I am seeing it."2209

There are some really cool ideas in math, and lots of the cool ideas in math end up taking a little while to fully understand.2215

So, even things that you don't understand the first time--the second time you look at it, it might make a lot more sense.2220

I think this stuff is really cool.2224

All right, we are ready for our final example.2225

For each limit below, if it exists, determine the value using the associated graph.2228

Our first one is limit as x goes to 0 of 2x/x; well, that one is pretty easy.2232

It is clearly going towards 2, because it is just in a steady state at 2 all the time.2237

While actually plugging in 0...if we plug in 0, 2(0)/0 gets us 0/0; we can't plug in 0.2242

So, it doesn't exist at 0/0; but on the way to that x = 0, it exists just fine.2248

We can see what the limit that it is going towards is.2254

The limit that it is going towards is 2.2257

Similarly, over here, the limit as x goes to -2 of (-3x - 6)/(x + 2): what is it going towards?2260

It is going towards that thing, which is -3; so it is going to -3.2267

If we actually plugged -2 in, well, x + 2, so -2 + 2, would get us dividing by 0; things would fall apart; we are not allowed to do that.2272

But as long as we are not plugging in -2 directly, everything is fine.2286

We always end up having -3, as we can see from this picture right here.2291

So, since it is always -3, then the limit of what it is going towards (we don't have to worry about that thing that it is actually at) is -3--as simple as that.2295

Also, I want to point out to you something that we are going to see in the finding of the values of limits.2305

We can actually get an early start on what is going to happen, to get that idea percolating through your head.2309

Look: 2x/x...well, at x = 0, it ends up doing something different.2313

But with the exception of x = 0, 2x/x behaves exactly like 2.2319

So, because it behaves exactly like 2, we can effectively say, "Well, what would it be at 0, if we were using this other alternative way of talking about it?"2325

The other alternative is that you are always 2; so since it is always 2, we end up getting a limit of 2 out of it.2334

The same thing is going on over here with (-3x - 6)/(x + 2).2340

Well, we can rewrite that top as -3 times (x + 2) over (x + 2), which means we can then write an equivalent thing,2349

with the exception of x = -2; everywhere else, we won't have that issue.2358

It would just be the same as -3 forever and always.2363

So, with the exception of that x = -2, it works just fine.2367

But we don't actually care about -2; remember, there is that idea of blacking out, where you are covering up that chunk,2370

because we are not allowed to peek underneath the cover.2379

So, if we cover up that chunk, and then we try to figure out where it is headed towards,2382

we don't have to worry about the fact that, if we had plugged in -2 here, it wouldn't work,2388

because we are not worried about what happens at -2.2392

We are just worried about what happens everywhere else; and what it is equivalent to everywhere else is -3.2394

And so, that is why we end up seeing it.2400

This is just a quick preview of the ideas we will end up getting when we start talking about finding limits precisely with more algebraic methods.2401

All right, that gets us a really good idea of how limits work.2407

Just remember: it is the idea of where you are going and if it matches from both sides;2410

the question is of where you are going, but it doesn't matter what it actually is there.2414

It is about the journey, not the destination.2417

All right, we will see you at Educator.com later--goodbye!2420