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Finding Limits
 The easiest limits to find are the limits of "normal" functions. Now "normal" is not a technical term, it's just supposed to mean the kind of functions we're used to dealing with. Functions where
 It does not "break" at the point we're interested in (the function is defined and makes sense);
 It is not piecewise and/or the point we're interested in is not at the very edge of the domain.
 The above is true because, in general, most of the functions we're used to working with don't do anything "weird". That is, they're defined everywhere, they don't have holes, and they don't jump around. Therefore the functions we're used to working with go where we expect them to go.
This is true even if f(x) has "weird stuff" happening somewhere else. All we care about is x→ c, so as long as the neighborhood around x=c is normal, this works.If f(x) is "normal" around x=c, then
lim
f(x) = f(c).  Often we can't use the above because something "weird" does happen at x=c. A common weird thing is dividing by 0. In this case, we can sometimes find the value of the limit (if it exists) by canceling factors before taking the limit.
lim
KA KB
=
lim
A B
 For a radical expression (one with a root), the conjugate is the same expression, but flipping the sign on one side:
√x^{2} − 3x
− 47x
⇔
√x^{2} − 3x
+ 47x  Conjugates allow us to rationalize fractions, which can sometimes help us find the value of a limit. By multiplying the top and bottom of a fraction by the appropriate conjugate, we can often find limits that would not otherwise be possible.
 We'll discuss evaluating the limits of piecewise functions in the next lesson, Continuity and OneSided Limits. For now though, remember: as long as you're not trying to evaluate a limit on a piecewise "breakover" (where it switches from one function piece to another), the function is probably behaving "normally" on the pieces that contain the point you care about. Thus, if it's not a breakover point, you can approach it like you're evaluating the limit of "normal" function: just plug in the appropriate value and see what comes out.
Finding Limits
 Notice that the function in the limit (2x^{2} + 4x −5) is a "normal" function. There's nothing strange or weird about it: it's just a standard polynomial. It doesn't "break" anywhere, it doesn't jump around, and it just generally works as we would expect it to work.
 When dealing with a "normal" function (or, if the whole function is not "normal", then the neighborhood around the location we're interested in is), we can almost always just plug in the location that limit is approaching. That is, if the function is "normal" at the location being approached, then
lim
x→ cf(x) = f(c).  As we noticed in the first step, f(x) = 2x^{2} + 4x −5 is a "normal" function. That means we can find the limit by simply plugging in the location it is headed towards: x→ −3, so plug in c=−3:
lim
x→ −32x^{2} + 4x −5 = 2(−3)^{2} + 4(−3) −5 = 2·9 −12 −5 = 18 −17 = 1

 

 


 As we noted in the previous problem and in the lesson, when we're dealing with a "normal" function, we can almost always just plug in the location that the limit is approaching. This can still be true even if the function we're working with is not "normal": all we need is that the neighborhood around the location behaves in a "normal" fashion. Thus, if the function is "normal" at the location being approached, then
lim
x→ cf(x) = f(c).  Notice that f(x) is decidedly not "normal": it is a piecewise function, so it does something "weird"it jumps whenever it switches pieces. The function has "breakover" points at x=2 and x=4: these make the boundary edges between the different pieces of the function. However, we also must notice that we aren't interested in those "breakover" locations! We're interested in x→ 5, so there is a small neighborhood of "normalcy" around c=5. Thus, even though f(x) is not "normal" in general, we can still plug in because the specific location is in a "normal" section.
 Because f(x) is "normal" in the neighborhood of c=5 (the location of the limit is not a "breakover"), we can plug in. [When evaluating the function, make sure to use the appropriate piece from the piecewise function.]
lim
x→ 5f(x) = f(5) = −2(5)+7 = −10+7 = 3
 Begin by noticing that we can not simply plug c=4 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
 Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. Factor the top and bottom of the fraction as completely as possible. This problem is fairly simple, and it's not too hard to see that a factor of (x−4) is in the top as well:
lim
x → 43x−12 x−4=
lim
x → 43(x−4) x−4  While we can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. (After all, at this point there isn't even an x variable remaining, so the limit has no effect: the limit of a constant is just the constant.)
lim
x → 43x−12 x−4=
lim
x → 43(x−4) x−4=
lim
x→ 43
Thus, by canceling out the common factor, we were able to make the function "normal" at the limit location and find the limit by plugging in.
lim
x→ 43 = 3
 Begin by noticing that we can not simply plug c=−6 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
 Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. Factor the top and bottom of the fraction as completely as possible.
lim
x → −612x+72 x^{2}−36=
lim
x → −612(x+6) (x+6)(x−6)  While we can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is x → −6, so plug in c=−6:
lim
x → −612x+72 x^{2}−36=
lim
x → −612(x+6) (x+6)(x−6)=
lim
x → −612 (x−6)
Thus, by canceling out the common factor, we were able to make the function "normal" at the limit location and find the limit by plugging in.
lim
x → −612 (x−6)= 12 (−6−6)= 12 −12= −1
 First off, don't get freaked out by the fact that the limit is based around a instead of x. The specific variable being used is unimportant: everything still works the same way. Next, notice that we can not simply plug c=3 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
 Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. Factor the top and bottom of the fraction as completely as possible. Factoring the top might be a little challenging, but notice that we can rewrite a^{4} as (a^{2})^{2}. With this in mind, we get
lim
a → 3a^{4}−81 a^{2}+2a−15=
lim
a → 3(a^{2}−9)(a^{2}+9) (a−3)(a+5)=
lim
a → 3(a−3)(a+3)(a^{2}+9) (a−3)(a+5)  While we can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is a → 3, so plug in c=3:
lim
a → 3a^{4}−81 a^{2}+2a−15=
lim
a → 3(a−3)(a+3)(a^{2}+9) (a−3)(a+5)=
lim
a → 3(a+3)(a^{2}+9) (a+5)
Simplify from there:
lim
a → 3(a+3)(a^{2}+9) (a+5)= (3+3)(3^{2}+9) (3+5)(3+3)(3^{2}+9) (3+5)= (6)(9+9) 8= (6)(18) 8= 3 ·9 2= 27 2
 Begin by noticing that we can not simply plug c=0 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
 Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function involves square roots, though: this brings up the idea of rationalization. Hopefully, if we rationalize the numerator of the fraction, we will notice a common factor that can be canceled. Let's try that. [Remember, you rationalize by multiplying top and bottom by the conjugatethe same expression, but with a flipped sign.]
While we could distribute the x in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:
lim
x → 0
√2+x−√2 x·
√2+x+√2
√2+x+√2 =
lim
x → 0(2+x)−2 x(
√2+x+√2)
lim
x → 0(2+x)−2 x(
√2+x+√2) =
lim
x → 0x x(
√2+x+√2)  While we still can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is x → 0, so plug in c=0:
lim
x → 0x x(
√2+x+√2) =
lim
x → 01 (
√2+x+√2)
Thus, by rationalizing to find a common factor and then canceling it, we were able to make the function "normal" at the limit location and find the limit by plugging in.
lim
x → 01
√2+x+√2 = 1
√2+0+√2 = 1 √2+√2= 1 2√2
 First off, don't get freaked out by the fact that the limit is based around k instead of x. The specific variable being used is unimportant: everything still works the same way. Next, notice that we can not simply plug c=−4 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
 Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function involves square roots, though: this brings up the idea of rationalization. Hopefully, if we rationalize the numerator of the fraction, we will notice a common factor that can be canceled. Let's try that. [Remember, you rationalize by multiplying top and bottom by the conjugatethe same expression, but with a flipped sign.]
While we could expand (FOIL) the two factors in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:
lim
k → −4
√k+13−3 k+4·
√k+13+3
√k+13+3 =
lim
k → −4(k+13) − 9 (k+4)(
√k+13+3)
lim
k → −4(k+13) − 9 (k+4)(
√k+13+3) =
lim
k → −4k+4 (k+4)(
√k+13+3)  While we still can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is k → −4, so plug in c=−4:
lim
k → −4k+4 (k+4)(
√k+13+3) =
lim
k → −41 (
√k+13+3)
Thus, by rationalizing to find a common factor and then canceling it, we were able to make the function "normal" at the limit location and find the limit by plugging in.
lim
k → −41
√k+13+3 = 1
√−4+13+3 = 1 √9 + 3= 1 3+3= 1 6
 Begin by noticing that the function, while unusual, is not actually "weird". While it's not something we're used to working with, it never breaks down. Why? The numerator and denominator are both defined for any x value, and the denominator can never equal 0 (because although e^{x} can get close to 0 as x → −∞, it can never actually reach 0). Thus, we see that the function is actually "normal".
 When dealing with a "normal" function (or, if the whole function is not "normal", then the neighborhood around the location we're interested in is), we can almost always just plug in the location that limit is approaching. That is, if the function is "normal" at the location being approached, then
lim
x→ cf(x) = f(c).  Thus, since the function we're dealing with is actually a "normal" function (even if it's one we're not used to working with), we can just plug in c=0:
From here, just simplify. [Remember that cos(0)=1 (from the unit circle) and e^{0}=1 (because anything raised to the 0 equals 1).]
lim
x→ 0cos(x) e^{x}= cos(0) e^{0}cos(0) e^{0}= 1 1= 1
 First off, don't get freaked out by the fact that the limit is based around v instead of x. The specific variable being used is unimportant: everything still works the same way. Next, notice that we can not simply plug c=0 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
 Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function currently has fractions inside of fractions, though: this brings up the idea of simplifying it so no fractions appear in the numerator or denominator. Hopefully, if we find a way to express the function without all these extra fractions, we will notice a common factor that can be canceled. Let's try that. [Remember, we can knock out these fractions by multiplying the top and bottom by each denominator.]
While we could distribute and expand all the factors in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:
lim
v→ 01 v−2+ 1 2v· (v−2)(2) (v−2)(2)=
lim
v→ 02 + (v−2) v(v−2)(2)
lim
v→ 02 + (v−2) v(v−2)(2)=
lim
v→ 0v v(v−2)(2)  While we still can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is v → 0, so plug in c=0:
lim
v→ 0v v(v−2)(2)=
lim
v→ 01 (v−2)(2)
Thus, by "cleaning out" the nested fractions to find a common factor and then canceling it, we were able to make the function "normal" at the limit location and find the limit by plugging in.
lim
v→ 01 (v−2)(2)= 1 (0−2)(2)= 1 (−2)(2)= − 1 4
 First, notice that while we can't really work with [(f(x+h) − f(x))/h], we can evaluate the function with those inputs to give us something we can work with:
Second, notice that we have h→ 0, not x → 0. This means we are looking for a way to swap out h for 0. However, if we try to plug in 0 for h right now, it will break down because we'll have a denominator of 0.
lim
h→ 0f(x+h) − f(x) h=
lim
h→ 0
√x+h− √x h  Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function involves square roots, though: this brings up the idea of rationalization. Hopefully, if we rationalize the numerator of the fraction, we will notice a common factor that can be canceled. Let's try that. [Remember, you rationalize by multiplying top and bottom by the conjugatethe same expression, but with a flipped sign.]
While we could distribute and expand the factors in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:
lim
h→ 0
√x+h− √x h·
√x+h+√x
√x+h+√x =
lim
h→ 0(x+h) − x h(
√x+h+√x)
lim
h→ 0(x+h) − x h(
√x+h+√x) =
lim
h→ 0h h(
√x+h+√x)  While we still can't just plug in the location of the limit, we can now cancel out the common factor:
After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is h → 0, so plug in c=0:
lim
h→ 0h h(
√x+h+√x) =
lim
h→ 01 (
√x+h+√x)
Finally, simplify the expression as much as you can. It won't come out as a number, but that's okay: we weren't told a value to use for x, so x should appear in the final result.
lim
h→ 01
√x+h+√x = 1
√x+0+√x 1 √x+√x= 1 2√x
*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
Finding Limits
Lecture Slides are screencaptured 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
 Introduction
 Method  'Normal' Functions
 Method  'Normal' Functions, Example
 Method  'Normal' Functions, cont.
 The Functions We're Used to Working With Go Where We Expect Them To Go
 A Limit is About Figuring Out Where a Function is 'Headed'
 Method  Canceling Factors
 One Weird Thing That Often Happens is Dividing By 0
 Method  Canceling Factors, cont.
 Notice That The Two Functions Are Identical With the Exception of x=0
 Method  Canceling Factors, cont.
 Example
 Method  Rationalization
 Method  Piecewise
 Example 1
 Example 2
 Example 3
 Example 4
 Example 5
 Example 6
 Intro 0:00
 Introduction 0:08
 Method  'Normal' Functions 2:04
 The Easiest Limits to Find
 It Does Not 'Break'
 It Is Not Piecewise
 Method  'Normal' Functions, Example 3:38
 Method  'Normal' Functions, cont. 4:54
 The Functions We're Used to Working With Go Where We Expect Them To Go
 A Limit is About Figuring Out Where a Function is 'Headed'
 Method  Canceling Factors 7:18
 One Weird Thing That Often Happens is Dividing By 0
 Method  Canceling Factors, cont.
 Notice That The Two Functions Are Identical With the Exception of x=0
 Method  Canceling Factors, cont.
 Example
 Method  Rationalization 12:04
 Rationalizing a Portion of Some Fraction
 Conjugate
 Method  Rationalization, cont.
 Example
 Method  Piecewise 16:28
 The Limits of Piecewise Functions
 Example 1 17:42
 Example 2 18:44
 Example 3 20:20
 Example 4 22:24
 Example 5 24:24
 Example 6 27:12
Precalculus with Limits Online Course
Transcription: Finding Limits
Hiwelcome back to Educator.com.0000
Today, we are going to talk about finding limits.0002
We often need to find the precise value that a limit will produce.0004
However, the methods we saw when we first introduced limits (that is, graphing it or a table of values for the function) are not precise.0007
They can give us a good idea of what the limit will be, but they don't give us certainty.0015
They don't let us know that it will be exactly something.0020
Likewise, the formal (ε,δ) definition of a limit that we talked about in the last lesson0022
and it is totally fine if you do not know it; that was a completely optional lesson, only if you are really interested in math,0027
and wanted to find out more about stuff that is going to come later in a few years if you keep studying math;0033
it is totally fine if you didn't do itbut if you did, that is still not really going to help us find limits.0038
It allows us to formally prove that this limit has to be here.0043
And it is the deeper mechanics of what is going on "under the hood" for how a limit works.0047
But it doesn't let us find limits; it doesn't make finding them easier; it is just about proving limits.0051
In this lesson, we will see various methods to find the precise value; that is what this lesson will be about.0056
The basic idea that we are going to see with all of these methods is to transform the function into something0061
that works pretty much the exact same way, that we will just be able to plug in a value for the x0066
in this equivalent version, and we will be able to churn out some value for what the limit will come out to be.0072
Now, before you watch this, make sure that you are already familiar with the concept of a limit.0077
You really want to have a good understanding of how a limit works.0083
We will be working out how to get numbers in this lesson.0086
But if you don't actually understand what this stuff means, it is all going to fall apart really quickly.0088
So, it is really important that you understand how a limit works before you watch this.0092
If you don't already have a good understanding of how limits work, check out the lesson two lessons back,0096
Idea of a Limit, where we will explain and get an idea of what a limit is about.0101
And that way, we will have some meaning for how we actually get precise values.0105
You can figure out the precise values without really understanding what is going on.0110
But that will fall apart really, really quickly; and you might as well just have a nice foundation to work from there.0114
It won't take that long to get an idea of what is going on.0119
All right, first, the easiest limits to find are limits for "normal functions."0121
And now, that is in quotes, because normal is not a technical term.0127
What I just mean here is...it is supposed to mean the kind of functions that we are used to dealing with,0131
the sort of thing that we use most often, functions where it does not break at the point we are interested in0135
that is to say, the function is defined and makes sense; it doesn't suddenly break down when we get to the place that we really care about.0142
And it is not piecewise, and/or the point that we are interested in is not at the very edge of the domain.0148
The point that we care about, whatever x we are going to (we are going to some x going to c)...0154
whatever c we are going towards, it is not going to be at the very edge of the domain, or where we split on some piecewise function.0159
So, as long as that location makes senseeverything in that area makes sense0166
we know how to use the function in that area around that place, and it isn't piecewise0169
there aren't different parts of it, and it is not the very edge of the domain,0174
the very starting or very last value for itas long as those aren't the case, we will be able to do it really easily.0177
If these two conditions are met, where it isn't breaking down and it is not piecewise,0184
and the point isn't at the edge of the domain, then it is really easy to figure out what the value is going to be.0189
So, if the point we are interested in is the value that x approaches in the limit0195
that is to say, x goes to c, and c is the place where things aren't breaking down,0200
and c is not at a piecewise breakover, and it is not at the very edge of the domain,0204
then it is almost always going to be the case that the limit as x goes to c of f(x)0208
is as simple as just plugging c in for x and getting f(c).0212
So, let's get an example first, to see how we could use this.0218
If we looked at the limit as x goes to 2 of 1/x^{2}, what would that end up being?0220
Well, first notice: while 1/x^{2} breaks down (it isn't defined, that is to say) at x = 0, that is here.0225
But we don't care about x = 0; that is not the area that we are interested in.0232
We care about x going to 2; so if x is going to 2, if we look in that area over here, that region, that is totally fine; it makes perfect sense.0236
1/x^{2} works fine in the region around x going to 2.0245
As x goes to 2, well, all of this stuff makes sense; we can see it.0250
It clearly just maps to values, and it is perfectly reasonable.0254
Second, 1/x^{2} is not a piecewise function; we don't have to worry about that.0258
And x = 2 is not at the edge of the domain (and the domain for 1/x^{2} is every x, with the exception of x equaling 0; it is all x not equal to 0).0261
So, we are not at the edge of a domain; it is not piecewise; and it makes sense in the area we are looking for.0270
It makes total sense in here; so since it does all of those, it means that we can just plug in the value that we are going to.0275
Because of these two conditions, limit as x goes to 2, we just plug it in for x,0282
and we have 1/2^{2}, which simplifies to 1/4, and there is our limit.0286
Why can we do thiswhat is the reason that we can get away with doing this?0291
Well, in general, most of the functions we are used to working with don't do anything weird.0294
They aren't strange in any way; and what I mean by not being weird is that they are defined everywhere; they don't have holes; they don't jump around.0299
They are defined anywhere that we might be interested in looking; they don't have any holes in them; and they don't jump around.0311
They work in a pretty reasonable way; they work normallythey are not weird.0317
What this all means is that the functions we are used to working with, the functions that we normally work with, go where we expect them to go.0322
Since a limit is about figuring out where a function is headed (a limit is about what our expectation is0330
for this function), and a normal function has our expectations fulfilled (what we expect from the function0336
is what we get out of the function), that means that we can evaluate normal functions at the location the limit approaches.0343
Our expectation, the limit as x goes to c of f(x), what we expect to end up landing on in our journey,0349
what we expect as we come in, ends up being what it actually iswhat it is at the location.0356
So, if it is normal, if f(x) isn't doing something weird, then what we expect ends up being what we actually get.0363
So, the limit as x goes to c...well, we can just plug that in for f(x),0371
and we have that f(c) will end up being the limit, any time that we are dealing with a normal function.0374
In fact, this is true even if f(x) does have weird stuff, but as long as it happens somewhere else.0381
All we care about is x going to c; so as long as the neighborhood around x going to c is normal,0387
and the weird stuff happens off somewhere elseit isn't happening directly on top of that c0392
then the weird stuff...we don't care about it, because the region we care about being normal is the region around x = c.0397
As long as the neighborhood around x = c is normal, as long as around x = c does this,0403
then we will end up being able to just plug into that, just fine.0409
So, if there is weird stuff, that can be OK, as long as it is far enough away.0413
When we looked at 1/x^{2}, there was weird stuff at x = 0, but we didn't care about x = 0.0416
We cared about x going to 2; so as long as the weird stuff isn't right on top of where we are going to,0421
we can just plug in our c, and that will tell us what the limit will come out to be,0426
if it is this fairly normal function that we have been working with for years and years.0431
Of course, sometimes x goes to c, but something weird does happen at x going to c.0436
When we are at x = c, it is a weird place.0441
One weird thing that often happens is dividing by 0; you are not defined when you divide by 0.0445
So, consider the function and limit below: f(x) = 2x/x.0450
Well, we can see it graphed here; it makes perfect sense, up until we try to plug in x = 0, at which point the function breaks down.0455
But the limit is pretty clear; it is going to 2 the entire time, so what is it headed towards?0462
It is headed towards 2; that is what we get out of the limit.0466
So, of course, we can clearly see that the limit in this case exists, and it is 2.0469
However, let's also notice that with the exception of x = 0, everywhere other than actually at x = 0,0474
where the weird thing happens, the function f(x) = 2x/x is just equivalent to if we had canceled out those x's and gotten g(x) = 2.0481
Notice: f(x) = 2x/x and g(x) = 2...these two functions are identical, with the exception at x = 0.0492
Everywhere other than x = 0, these two are totally the same.0503
f(x) = 2x/x and g(x) = 2 behave exactly the same, with an exception at x = 0.0509
However, since we are looking at the limit as x goes to 0, we don't care about x = 0.0516
It is about the journey, not the destination; so if it is x to 0, the destination we don't care about is at x = 0.0523
So, the weird thing here at x = 0...we might as well forget about it.0529
That means, since we don't care about what happens at the weird place, and g(x) = 2 is exactly the same0533
everywhere but the weird place, that means that we can use g(x) to evaluate the limit.0540
And we can just plug in g(0) to get 2.0545
So, g in general works just the same as f; g is just the same as f.0548
It works the same everywhere, with the exception of this one weird little point.0555
However, since we are looking at a limit going to that one weird little point, we don't actually care about the weird little point.0559
A single point doesn't matter, because the limit is about the journey towards that point.0564
So, g(x) = 2 behaves the exact same for the journey portion.0568
The journey towards behaves the exact same, whether we are using f or g, which means that g(x) is what we can use for figuring out the limit.0574
And then, by the same logic we were just talking about previously, since g(x) is totally normal at 2,0584
we can just plug in there, and we can get the answer from g(x).0588
Now, how did we find g(x)?by canceling common factors.0591
This logic works in general; if we have some function given as a fraction, we can cancel out factors between top and bottom,0598
because a single point...if we cancel a factor, the only thing that can possibly happen0607
that would be bad is that we would cause one point to change around, like with f(x), x = 0 did not exist;0611
but with g(x), x = 0 did exist; so we caused a problem for specifically one point.0617
But a single point has no effect on a limit, because with a limit, it is about the journey, not a single point that is our destination.0622
So, since a journey is made up of a multitude of points, taking out a single point,0634
changing a single point, doesn't actually have an effect on where we end up landing for our limit.0638
That means, any time we have common factors for a limit, we can cancel out common factors0643
and just get what it would be without those common factors for the limit.0647
We will have changed the function that we are using, but the limits will be equivalent.0651
So, here is an example: we have limit as x goes to 3 of (x  3)(x + 2)/(x  3).0656
Well, that means that, if we were to plug x = 3 into this, we would have 0/0; 3  3 turns to 0; 3  3 on the bottom turns to 0; so we would get 0/0.0661
And so, we can't do that; it goes crazy; it is weird there.0671
But we can cancel out x  3 and cancel out x  3, just like we canceled out the k's here; and we get some other something over something.0674
We get what remained, a and b; in this case, we have x + 2 divided by 1 now.0682
So now it is the limit as x goes to 3 of x + 2; that is effectively going to work the same.0687
x + 2 is pretty much equivalent to x  3 times x + 2 over x  3, with the exception of x = 3.0692
But since we don't care about that, since that is where we are headed towards, we can end up using that limit instead.0700
So, we now plug in x going to 3 into x + 2.0704
Well, x + 2 is perfectly normal at x = 3; nothing weird happens there.0707
So, since nothing weird happens there, it works normally; we can just plug our value in there.0712
We plug in 3; 3 + 2...we get 5; our answer is 5 for the limit.0716
A similar idea to canceling out factors is rationalization: rationalizing a portion of some fraction.0722
If we have a radical that is in our way, and it is part of a fraction, we can change it into a nonradical0728
by multiplying the numerator and denominator by the same thing.0735
What do we get rid of a radical withhow do we rationalize an expression?0738
We rationalize an expression that contains a radical by multiplying by its conjugate.0743
That is the same expression, but now with a negative on one side.0748
For example, if we have √x + 2, well, that is a radical and some nonradical thing.0752
If we go through the conjugate process, we get √  2; plus switches to negative.0756
If we have √(x^{2}  3x)  47x, some radical minus something that is not in a radical,0762
then its conjugate is √(x^{2}  3x) + 47x.0768
So notice: plus, if we are going through a conjugate, becomes minus; and minus becomes plus.0773
We just swap the sign on one of the portions, and that is how we get the conjugate.0781
Now, if we multiply a radical expression by its conjugate, we will end up canceling out the radicals.0785
And we will see how that works in just a moment.0790
Since multiplying a fraction on the top and bottom by the same thing gives an equivalent expression0793
if I have a fraction, I can multiply it by 5/5, because 5/5 is just the same thing as 1, so it is still equivalent0798
we can figure out limits by doing this thing.0804
And we can trust that this works, that this doesn't introduce any issues, by the exact same logic that we use to cancel out factors.0806
If we were to multiply by something over something, it could introduce one slight issue;0813
but it would only introduce a slight change to the function at one point.0818
But because it is a limit, we don't care about single points on their own; we only care about continuums of points.0821
So, that single point being changed is not really an issue; the limit will still work the same.0826
So, if we have the limit as √(x + 4)  2, over x, as x goes to 0,0831
well, we would want to multiply this by the conjugate, √(x + 4)...but it was a minus previously,0838
so now it swaps to a + 2, and it will have to be divided by the same thing,0844
because we can only multiply by 1 effectively, which means the same thing on top and bottom: + 2.0848
And we multiply by that, and we start working things out.0854
What do we get out of that? Well, limit as x goes to 0 of √(x + 4)  2, over x, times √(x + 4) + 2,0857
over √(x + 4) + 2...well, the top here is now going to become x + 4  4.0864
And if we are not quite sure how we see that, let's look at √(x + 4)  2 times √(x + 4)  2.0869
Well, what is...oops, I should not have put that radical over the whole thing; the radical ends there.0878
What is √(x + 4) times √(x + 4)?0883
Well, the square root of thing times the square root of same thing always just lets out the thing on its own.0885
The square root of smileyface times the square root of smileyface becomes smileyface.0890
So, √(x + 4) times √(x + 4) becomes x + 4.0894
So, x + 4 is what we get out of that; and now we have √(x + 4) times 2...0898
oops, that shouldn't be 2; it should be plus, because it is a conjugate; I'm sorry about that.0905
√(x + 4) times positive 2 is 2√(x + 4): and then 2 times √(x + 4) is 2√(x + 4).0909
So, we have positive 2 √(x + 4) and 2√(x + 4); those two things cancel each other out.0916
Positive 2 and negative 2 cancel each other out; so the middle part disappears.0922
And now, it is 2 times +2; that becomes 4; so that is where we get x + 4  4.0927
On the bottom, it is x times this thing over here; so we just put in the quantity, because we will have some convenient canceling happening very soon.0934
On our top, we have x + 4  4; so plus 4 and minus 4 cancel each other out; we are left with just x on top;0943
it is still x times √(x + 4) + 2 on the bottom.0948
But now, we say, "We have x on top and x on bottom," and now we have the limit as x goes to 0 of 1/(radic;(x + 4) + 2).0952
At this point, we see that if we plug in 0...does anything weird happen?0959
Well, 1/(√(0 + 4) + 2)...we aren't dividing by 0 anymore, so it effectively works as a normal function.0964
We don't have any weird thing happening, so we plug in for our x at this point.0971
The square root of 0 + 4, plus 2 is in our denominator: 1 over...√4 is 2, so we have 1 over 2 + 2, which gets us 1/4.0975
And that is how we get to our answer for that limit.0985
We will discuss evaluating the limits of piecewise functions.0988
We haven't talked about piecewise functions yet, because we will be talking about that in the next lesson, Continuity and OneSided Limits.0991
We will want a couple of little new ideas before we talk about piecewise functions.0997
That is why we are saving them for the next lesson.1001
For now, though, remember: as long as you are not trying to evaluateas long as it is not trying1002
to evaluate the limit at some piecewise breakover, where it switches from one piece to another piece,1007
the function is probably going to be behaving normally on the pieces that contain the point you care about.1014
It might have a piecewise here and a piecewise here, and then it suddenly switches over.1018
But as long as we are over here in the normal area, or we are over here in the normal area, nothing weird happens.1021
So, if it is not at a breakover point, then that means that, since it is behaving normally,1027
you can approach it the same way as you do when just dealing with the limit of a normal function.1032
Just plug in the appropriate value and see what comes out.1037
Plug in the value that you are going towards and see what comes out.1040
However, if you do need to evaluate a breakover point, where you have to be talking1043
about where it switches from one to the next, check out the next lesson,1048
because we will see how that idea works specifically in the next lesson.1051
All right, we are ready for some examples.1055
The first one: Evaluate the limit as x goes to 2 of x^{4}  3x^{2} + 4x  10.1057
Our first question that we want to ask ourselves is if x^{4}  3x^{2} + 4x  10 is normal.1062
Yes, there is nothing weird that happened in that; that is just a polynomialwe are used to that sort of thing.1066
So, it is normal; if it is normal, then that means that we can just plug in our value for each of the x.1071
So, we plug in our 2, because we know that the limit of what it gets is...well, we don't have to plug x into 10...1078
we know that the limit as what we are going to get out of this is just the same thing as what the function would be there.1084
What we expect is what we get when we are dealing with a normal function.1089
So, we plug in 2; we now have 2^{4}  3(2)^{2} + 4(2)  10.1094
2^{4} is 16; minus...3(2)^{2}...2^{2} is 4; 3 times 4 is 12,1103
plus...4 times 2 is 8; minus 10; 16 + 8 gets us 24; minus 12 minus 10 gets us 22; we put those together, and we now have 2done!1109
The next one: let f(x) equal x^{2}  3 when x is less than or equal to 2 and 5x + 2 when x is greater than 2.1120
First, let's just see, really quickly, what this looks like.1127
Here is a rough picture of what it looks like; I will do that with blue here; so x^{2}  3...that is basically like a parabola.1131
It has just been lowered by 3 from a normal standard parabola.1139
And it goes until x ≤ 2, at which point it stops here, at 2.1143
And then, after that, we are at x > 2, so we switch over to 5x + 2 when x is greater than 2.1147
It starts here, and then it goes off like this.1153
And that is what we end up seeing.1155
The question here is: if we are going to evaluate the limit as x goes to 1 of f(x)...oh, no, it is a piecewise function!1157
Well, yes; but we are clearly contained within x ≤ 2.1165
The area we care about is this area here; that is far enough from something weird happening.1169
Something weird does end up happening over a little bit further to the right.1174
But we don't care about that, because in the neighborhood we are interested in (that is x going to 1),1178
we are definitely less than or equal to 2, if we are close enough to 1.1183
So, since being close enough to 1 means nothing weird happens, all we are dealing with is x^{2}  3.1187
So, we are effectively normal, because we only have to consider the part of the piecewise function that we are inside of.1194
The part of the piecewise function that we are inside of is x^{2}  3.1201
So, we can just plug our 1 into x^{2}  3; we plug in our 1; 1^{2}  3; 1  3 gets us 2, and there is our limit.1204
The next example: Evaluate the limit as x goes to 3 of (x^{2} + 3x)/(x^{2}  9).1216
Our first question is, "Is it normal?" Well, if we plug in 3, what do we get on the top?1221
Well, we will get 9  9; so that is 0; and on the bottom, we will get x^{2}  9,1226
so that will be positive 9, minus 9; oh, we get 0/0; so that means it is not normal.1231
Oh, no! But what do we do as soon as we are not normal?1237
We start thinking, "All right, well, what else can we do?"1241
We have the possibility of canceling factors; so what we want to do now is think, "Is there a way for us to cancel out factors?"1243
Can we cancel out factors? Well, we have x^{2} + 3x...1251
so I will write limit as x goes to 3...technically, we really should have the limit at every step.1254
If you end up really not feeling like writing out the whole thing, at least write limit, so that we know1259
that we are still dealing with the limit; and we will plug in something later.1263
(x^{2} + 3x)/(x^{2}  9): well, limit as x goes to 3...how can we change the top?1266
How can we factor the top? Well, we could pull out an x, and we would have x(x + 3) on top.1277
How can we factor the bottom, x^{2}  9? Oh, that is just the same thing as (x + 3)(x  3).1282
Great; we can cancel factors; we cancel x + 3 and cancel x + 3.1289
And that is fine, because we are just working with a limit.1295
Our limit is now the same thing as x goes to 3 of x on top, divided by x  3 on the bottom.1298
Now, we ask ourselves, "If we plug in 3, does anything weird and disastrous happen?"1307
3 on top; 3 on the bottom; we don't get 0/0; we don't even get dividing by 0 once.1311
We are totally fine; so we plug in, because now it is effectively a normal function, since something weird isn't happening.1317
So, we have 3/(3  3); that gets us 3/6; the negatives cancel; 3/6 is 1/2, and so we get 1/2 as the answer to our limit.1324
Great; the next one: Evaluate the limit as x goes to 0 of tan(x)/sin(x).1337
The first question that we ask ourselves is, "Is this normal?"1342
Well, if we plug in 0, tan(0) is 0; sin(0) is 0; oh, that means it is 0/0; so it is not normal.1347
But if it is not normal, the first thing we try is asking ourselves, "Can we cancel factors?"1357
So, if we can cancel factors, we are good; how can we cancel factors?1363
Limit as x goes to 0...how can we change tan(x)?1367
Well, tan(x)...remember, that is just the same thing as sin(x)/cos(x).1372
And any time we don't have just sine and cosine, and we are dealing with trigonometric stuff,1377
it normally helps to put it just in terms of sine and cosine; so we have sin(x)/cos(x), all divided by sin(x).1380
Oh, OK; well, now we see that we can start canceling stuff; limit as x goes to 0...1390
well, we could rewrite this as sin(x)/cos(x); we could just cancel out the sin(x) if we see that directly.1395
But if we find it difficult to divide fractions and fractions, we could think of this as divided by sin(x),1401
which means that that is the same thing as limit as x goes to 0 of sin(x)/cos(x), times 1/sin(x).1407
If you divide, it flips to a fraction, which we could have also done by just breaking out the fraction of the sin(x) on the bottom to the side.1416
So now, we see that there is sin(x) on the bottom and sin(x) on the top.1422
We cancel some stuff out; limit as x goes to 0...now we have 1/cos(x).1426
Since we managed to cancel some stuff out, let's ask ourselves, "If we plug in 0, is it weird? Is it normal? What happens?"1432
Well, x goes into 0 for cos(x); cos(0) is 1; 1/1 is totally not weird anymore.1438
So, that means we can now swap in our 0; 1/cos(0)...cos(0) is 1; we have 1/1, so our limit comes out to be 1.1445
OK, the next one: Evaluate the limit as x goes to 4 of (2  √x)/(4  x).1460
The first thing we ask ourselves is, "Is it normal?"1466
Well, if we plug in 4 on the top, we will get 2  √4, so that will come out to be 0,1468
divided by 4  4...even worse...dividing by 0...so 0/0 is not normalno! No!1472
It is not normal, so the next thing we ask ourselves is if we can cancel factors.1480
Well, 2  √x...4  x...we might be able to figure out a way to cancel factors, but not easily.1484
Canceling factors...we might be able to figure out a way; we could figure out a way;1490
but let's say that we don't want to figure out canceling factors; canceling factors is not easy.1496
So, there is a radical; what was the trick we learned for dealing with radicals?1501
We rationalize; we move on to the next trick in our selection; we rationalize.1505
What do we do? We have the limit as x goes to 4 of 2  √x, over 4  x.1514
How do we rationalize? We multiply by the conjugate on the top and the bottom.1523
2  √x: its conjugate is 2 + √x; we could also put a negative on the 2, but it doesn't really matter which side gets the negative.1526
2 + √x; 2 + √x; greatmultiply our tops together, and multiply our bottoms together.1534
2  √x times 2 + √x: these are now factors with parentheses around them, because we have to have distribution going on.1542
So, 2 times 2 is 4; 2 times √x is 2√x; √x times 2 is +2√x and 2√x; they cancel each other out;1550
√x times +√x becomes x (√x times √x always comes out to be x;1559
√smileyface times √smileyface always comes out to be smileyface; root(root) cancels the roots,1565
as long as it is the same thing underneath it).1569
4  x times 2 + √x: well, we could multiply them togetherbut 4  x is what we have on the top right now.1571
We are basically working towards canceling out factors; so 2 + √x...1577
at this point, we are now going to be able to cancel out factors.1582
We have the 4  x on the bottom and the 4  x on the top; they cancel each other out.1587
And now, I have the limit as x goes to 4 of 1/(2 + √x).1592
Great; so, if we were to plug 4 into this, would something weird happen?1601
1/(2 + √4)...no: we don't have 0 showing up; we aren't dividing by 0.1605
It basically works like a normal function; it does work like a normal function.1609
Nothing weird is going on there; so that means we can just plug in our value.1613
So, 1 over 2 plus...plugging in 4 for our x...1 over 2 + 2 (the square root of 4 is 2)...and 1/4 is our answer; nice.1617
The final example: Evaluate the limit as x goes to 0 of 1/(x + 3)  1/3, all divided by x.1632
The first question we ask ourselves is, "Is it a normal function?"1638
So, is it normal? Well...you can guess by my hint: no.1641
If we plugged in 0, it would be 1/3  1/3, so 0 on top, divided by 0 on bottom...no, it is definitely not normal.1646
So, it is not normal; oh, no, what are we going to do?1654
Well, the next thing we ask ourselves is, "Can we cancel factors?"1658
Can we cancel? Not easily: 1/(x + 3)  1/3...I really don't see any easy ways to make cancellation show up there.1663
So, we are probably not going to be able to cancel, at least not easily.1672
So, our next question is...last time we asked ourselves, "Can we rationalize?"1675
Well, there are no radicals here, so we can't rationalize.1678
But we can take a hint from the idea of rationalization.1681
The idea of rationalization was to multiply the top and the bottom by something that makes some part not weird anymore,1684
not as strange to deal with, so that hopefully, we can get cancellation to appear later.1689
Well, what would make the top...the thing that is really strange about this is that we have fraction over fraction.1693
We don't like fractions and fractions; so how can we get rid of some of those fractions by multiplying?1699
Well, the easiest way to get rid of the denominator in the top is to just multiply by the denominators in the top.1706
If we multiply...I will rewrite the thing out...limit as x goes to 0 of 1/(x + 3)  1/3, all over x:1712
well, what would get rid of the x + 3? x + 3 would get rid of the denominator of x + 3.1723
What would get rid of 1/3? Well, multiply by 3.1730
We can get rid of both of those denominators by multiplying that whole top by (x + 3) times 3.1733
That will cancel out each of the denominators as we work through it.1738
And remember: it is always going to multiply the whole thing; when we multiply, we multiply the quantity, because we have to have distribution showing up.1740
And on the bottom, we will have to have the same thing, because otherwise we are changing the expression; we can't change the expression.1747
x + 3, times 3, over x + 3, times 3; great; our limit continues...limit as x goes to 0...1754
What do we get on the top? Well, (x + 3) times 1/(x + 3)...those cancel out, and we are left with just the 3 left over.1760
So, (x + 3) times 3, on 1/(x + 3)...the (x + 3)s cancel out; we are left with just 3.1767
Minus...when (x + 3) times 3 hits 1/3, well, the (x + 3) doesn't do anything; but the times 3 cancels out, so we are left with minus (x + 3).1774
All right, now we could expand the bottom, but that won't actually help us.1785
One of the ideas that we are going to hopefully manage to get to is to figure out a way to cancel things.1789
We couldn't cancel things easily by factoring; but hopefully we will still manage to cancel something at some point later on.1793
We don't want to expand factors; we want to actually keep up this process of keeping things in factors.1799
So, let's not put anything together; we will have it as x times x + 3 times 3.1805
At this point, we see x + 3 on top and x + 3 on the bottom, but we have to be careful; don't cancel stuff.1810
We can't cancel, because there is still a subtraction sign on the top; we have to have the whole factor.1814
We keep working to simplify: limit as x goes to 0...3  (x + 3)...well, the 3's will cancel out,1818
and we will be left with just x on the top, divided by x times (x + 3) times 3; great.1825
Limit as x goes to 0 of x/x(x + 3)(3); at this point, we say, "We can cancel some stuff!"1838
This x and this x cancel, and we are left with the limit as x goes to 0 of 1 now (because it just canceled out the x,1844
not also the negative), times (x + 3), times 3.1852
Now, we ask ourselves, "Now that we have managed to cancel something, if we were to plug in a number, would we have something weird happen?"1856
Would it be normal now? If we plug in 0, we get 1/(0 + 3)(3); it doesn't look like we are going to be having dividingbyzero issues anymore.1862
It isn't weird anymore, so we can just plug in; this is equal to the 1, over (0 + 3) times 3.1869
Once it is not weird, we can plug in, because now it is effectively normal.1879
And when it is effectively a normal function, you can just plug into it with your limit.1882
1/3(3) gets us 1/9; and there is our answergreat.1887
All right, at this point, we have a really good understanding for how to figure out how limits work.1896
The basic idea is, "All right, I have a limit that is normal and doesn't have anything weird happen."1900
That is easyjust plug in something and crank it outsee what number you get out.1905
That is what the limit is, because "normal" means that your expectations will be met.1908
If it is not normal, if there is something weird happening, you try to manipulate things.1912
You either pull out factors or multiply the top and the bottom; you do something where you are allowed to cancel factors later on.1916
And then, you check and see, "OK, now that I have canceled out the factors, is it possible for me to plug things in and have it be normal, effectively?"1922
Can I now plug in (now that there isn't, hopefully, a weird thing happening)?1929
Sometimes there will still be weird things happening; in all of the examples we saw here, we canceled out anything that would cause weird stuff to happen.1932
But sometimes, you will end up still having weird stuff, no matter what you manage to cancel out.1939
And in that case, it can help to check a graph and think, "Oh, I see: it is going to go out to infinity," or something like that.1943
And you will see that it is never going to work.1947
But a lot of the time, you can cancel stuff out, and you can say, "Oh, OK, now it is effectively behaving like a normal function;1949
so I can plug in the xvalue that I am going towards and just crank out an answer."1954
All right, we will see you at Educator.com latergoodbye!1958
1 answer
Last reply by: Professor SelhorstJones
Thu May 9, 2013 4:50 PM
Post by Valentina Gomez on May 7, 2013
the example problems had me cracking up! Thank you for making limits seem easy!