Is Cady Right, or Does the Limit Exist?

I recently watched the movie Mean Girls. At the end of the movie, there is a scene where the main character is shown a limit and asked to evaluate it. After quickly working out the problem in a tight competition for the academic decathlon, Cady looks up from her paper and shouts, “THE LIMIT DOES NOT EXIST!”

And of course, being the math oriented person that I am, I just was left wondering whether the limit actually exists after Mean Girls concluded. So I figured I’d share my findings.

First of all, here’s the limit shown at the end of Mean Girls: $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)}$$

Where do we start?

Typically when I’m evaluating limits, I like to assume that the function is continuous at the point we’re evaluating the limit and just plug it in. This usually doesn’t work, but it will frequently give us a hint as to which method we can use to evaluate it. This basically just takes advantage of the basic limit properties of limits if it works out. But we need to make sure we’re not causing any problems when we do this.

So let’s just start with plugging in zero for x in the function who’s limit we’re looking for.

$$f(x) = \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)}$$ $$f(0)=\frac{ln(1-0) \ – sin(0)}{1-cos^2(0)}$$ $$f(0)=\frac{ln(1-0) \ – sin(0)}{1- \big( cos(0) \big)^2}$$ $$f(0)=\frac{0 \ – 0}{1- 1} = \frac{0}{0}$$

Turns out, this doesn’t really work this time. Because we end up with the indeterminate form of $$\frac{0}{0}$$. Since it’s never fine to divide by zero, this isn’t allowed.

So, we’ll need to think of another way to evaluate this limit. But like I said before, the fact that we got this indeterminate form gives us a hint of how we can evaluate this limit properly. This indeterminate form actually is one of the conditions that tells us we can evaluate this limit using L’Hospital’s Rule.

L’Hospital’s Rule

Which means we can create a new limit. We will still have $$x \to 0$$ in our new limit. But we’ll take the limit of a different function. And hopefully this one will be easier to evaluate.

This new limit will actually just be made up of a fraction where the top of the fraction is just the derivative of the original numerator. And the bottom of this new faction is just the derivative of the denominator of the original fraction. So we know that $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)} \ = \ \lim_{x \to 0} \ \frac{\frac{d}{dx} \big[ ln(1-x) \ – sin(x) \big] }{\frac{d}{dx} \big[ 1-cos^2(x) \big] }$$

Remember, we are not finding the derivative of the fraction as a whole, so we should not use the quotient rule. We will need to apply the chain rule to find the derivative of a few of these terms. Doing so tells us that $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)} \ = \ \lim_{x \to 0} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$

Now can we evaluate this limit?

The point of using L’Hospital’s Rule is that we should end up with a limit that’s easier to evaluate. So, let’s try the same thing we did earlier and see what happens. Since we have a limit as $$x \to 0$$, let’s just plug zero in for x and see what we get. $$\lim_{x \to 0} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ $$\frac{ – \ \frac{1}{1-(0)} \ – \ cos(0) }{ 2cos(0) \cdot sin(0)}$$ $$\frac{ – \ 1 \ – \ 1 }{ 2(1)(0)} = \frac{-2}{0}$$

And you can see we’ve divided by zero. So, this isn’t going to work. You can’t divide by zero, it breaks the rules of math. Therefore, we can’t just evaluate this limit by plugging in zero for x.

However, we are getting closer. The reason I say this is that we don’t have the indeterminate form of $$\frac{0}{0}$$ anymore. Since we have some other number divided by zero, that tells us that this limit will likely be either $$\infty$$ or $$– \infty$$.

The reason for this is that the numerator of the fraction is going toward the number -2. Whether you approach from the left or the right of x = 0, the top of our fraction is going to be a number close to -2.

Meanwhile, as x gets really, really close to 0, the denominator is going to also get really close to 0. But we don’t care about what the denominator is when x = 0. Since we’re dealing with a limit we care about the denominator when x is really close to 0. But depending on which side of zero we’re on, the denominator could be positive or negative.

If we have a fraction where the numerator is some non-zero number, and the denominator is getting infinitely close to zero, the fraction as a whole will become infinitely large. We just need to figure out if it’s positive or negative. Well, we can figure this out by splitting the limit up into two one-sided limits and see what those tell us about the two-sided limit.

Left-Sided Limit

Let’s start with the limit where x is approaching zero from the left. $$\lim_{x \to 0^-} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ We already know that the numerator of this fraction is going to approach -2 as x approaches 0 from either side. So, let’s just look at the denominator of this fraction.

What happens to $$2cos(x)sin(x)$$ as $$x \to 0$$? Well, we already said it will approach 0, but let’s just consider what’s going on as x approaches 0 from the left. If we’re approaching 0 from the left, that means we want to consider x values that are super close to 0, but will be negative. For negative x values approaching 0, cos(x) will approach 1, and will be slightly smaller than 1. And sin(x) will approach 0, and will be negative numbers very close to 0.

This is because sin(x) is negative for x values very close to 0. We can see this looking at a graph of $$f(x)=sin(x)$$.

Based on this, as x approaches 0, 2cos(x)sin(x) will become the product of 2, 1, and a negative number very close to 0. The product of two positive numbers and a negative number will be a negative number. So we know that this denominator will be getting closer and closer to 0, but will be a negative number in this left sided limit.

So we can think of the fraction as a whole, as a positive number very close to 2, divided by a negative number very close to 0. As x goes to 0 from the left, this fraction will get infinitely large and will be negative because it’s a positive divided by a negative. In other words, as $$x \to 0^-$$, $$\frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) } \to \frac{2}{-0}$$. Or $$\lim_{x \to 0^-} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }=-\infty$$

Right-Sided Limit

The right-sided limit will work out very similarly, but with one main difference. $$\lim_{x \to 0^+} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ Again, the numerator of this fraction is going to approach -2 as x approaches 0 from the right. But let’s look at the denominator.

By the same process as the left-sided limit above, we can see that the denominator of this fraction is going to approach 0. But what we need to figure out is whether it is going to be approaching 0 from the negative side or the positive side. And the only term of our denominator that is going to be different approaching from the right side of 0, is sin(x).

Looking back up at the graph of $$f(x)=sin(x)$$ you can see that the function is positive (above the x axis) to the right of $$x=0$$. Therefore, the denominator of this fraction is going to be the product of three positive numbers, resulting in a positive number. And the fraction as a whole will be a positive number over a positive number. In other words, as $$x \to 0^+$$, $$\frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) } \to \frac{2}{+0}$$. Or $$\lim_{x \to 0^+} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }=\infty$$

What does this tell us about the two-sided limit?

Remember, for a limit to exist both one-sided limits need to exist AND they both need to be equal to each other. We just figured out that our left-sided limit is $$-\infty$$ and the right-sided limit is $$\infty$$. Since one is positive and one is negative, they clearly aren’t equal. Each one-sided limit does not give the same result. Therefore, the limit we are trying to find indeed does not exist.

So, it turns out Cady was right in the end of Mean Girls. The limit does not exist.

Solution – Find the values of a and b that make the function differentiable everywhere.

Find all values of $$a$$ and $$b$$ that make the following function differentiable for all values of $$x$$.

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ ax+b & \mbox{if } x>-1 \end{cases}$$

When trying to solve a problem like this, there are actually two things you will need to consider for our function $$f(x)$$.  Obviously, we need to make sure that it’s differentiable everywhere, but this actually implies something else that we will want to consider as well.

Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous.  The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere.  Once we make sure it’s continuous, then we can worry about whether it’s also differentiable.

Making sure f(x) is continuous everywhere

I’m not going to go into quite as much detail to show the part about making sure the function is continuous because I have already done this, which you can see by clicking here.

To make sure $$f(x)$$ is continuous at $$x=-1$$ we need to make sure that $$\lim_{x \to -1} f(x) = f(-1).$$  Since we have a piecewise function, we will need to consider each one-sided limit, but in this case only the right sided limit will tell us something useful.

$$\lim_{x \to -1^{+}} f(x) = f(-1)$$

$$\lim_{x \to -1^{+}} ax+b = b(-1)^2-3$$

$$a(-1)+b=b-3$$

$$-a+b=b-3$$

$$-a=-3$$

$$a=3$$

So now we know that $$f(x)$$ will be continuous everywhere as long as $$a=3$$.  However, this doesn’t really tell us that $$f(x)$$ is differentiable everywhere as well.

Making sure f(x) is differentiable everywhere

We now know that we will need to let $$a=3$$ in order for this function to be continuous and to have a chance of being differentiable.  As a result, we can say that we are now trying to make this function differentiable everywhere:

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ 3x+b & \mbox{if } x>-1 \end{cases}$$

We can see that the only place this function would possibly not be differentiable would be at $$x=-1$$.  The reason for this is that each function that makes up this piecewise function is a polynomial and is therefore continuous and differentiable on its entire domain.  The only place we may have a problem is when we have to switch between the two functions.

What does it mean for a function to be differentiable?

It means that its derivative exists for all values of $$x$$.  In other words, we need to be able to find its derivative no matter what $$x$$ is.

However, as I mentioned above, in this case we really only need to make sure that we can find the derivative of $$f(x)$$ when $$x=-1$$ since we know it would exist for all other values of $$x$$.  By using the definition of a derivative, we need to make sure the following limit exists at $$x=-1$$.

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Since we need to check this when $$x=-1$$, we can plug in $$-1$$ for $$x$$.  Therefore, we need to make sure this limit exists:

$$\lim_{h \to 0} \frac{f(-1+h)-f(-1)}{h}$$

I went over this limit definition in greater detail previously.  If you want a refresher on where this is coming from you can find that by clicking here.

Just like when we had to find the limit to make sure that $$f(x)$$ was continuous, we will need to consider each one sided limit separately in order to find this limit.  And also like when we checked for continuity, each one sided limit is going to require the use of a different section of our piecewise function.

Setting up the limits

When $$h$$ is slightly less than $$0$$, and we are considering the left sided limit, $$f(-1+h)$$ would need to be found using the $$y=bx^2-3$$ because this would involve inputting $$x$$ values which are less than $$-1$$.

By the same reasoning, when $$h$$ is slightly greater than $$0$$, and we are considering the right sided limit, $$f(-1+h)$$ would need to be found using the $$y=3x+b$$ because this would involve inputting $$x$$ values which are greater than $$-1$$.  Therefore, we need to consider the following one sided limits:

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

Now what do we do with these limits?

Now remember, as I discussed in the lesson about one-sided limits, in order for a limit to exist we need both of its one-sided limits to exist and they need to be equal.  Therefore, in order to show that the derivative of $$f(x)$$ exists at $$x=-1$$, these two one-sided limits need to be equal to each other.  Before setting them equal to each other, first we’ll simplify them a bit.  First the left side limit.

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)(-1+h)-3\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(1-2h+h^2)-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b-2bh+bh^2-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{b-2bh+bh^2-3-b+3}{h}$$

$$=\lim_{h \to 0^{-}} \frac{bh^2-2bh}{h}$$

$$=\lim_{h \to 0^{-}} \frac{h(bh-2b)}{h}$$

$$=\lim_{h \to 0^{-}} bh-2b$$

$$=-2b$$

And now the right sided limit.

$$=\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{\Big[-3+3h+b\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{-3+3h+b-b+3}{h}$$

$$=\lim_{h \to 0^{+}} \frac{3h}{h}$$

$$=\lim_{h \to 0^{+}} 3$$

$$=3$$

Now if we set these two simplified versions of the one-sided limits equal to each other, we get

$$-2b=3$$

$$b=-\frac{3}{2}$$

What does this tell us?

So now if we put both pieces together, we know that $$a=3$$ will ensure that $$f(x)$$ is continuous and then making $$b=-\frac{3}{2}$$ will also make sure $$f(x)$$ is differentiable at $$x=-1$$.  This would in turn make $$f(x)$$ differentiable for all values of $$x$$, or make it differentiable everywhere.

As always, I want to hear your questions!  Go check out my other lessons about derivatives and if you can’t get your question answered, I’d love to hear from you.  Leave a comment below or email me at jakesmathlessons@gmail.com.  If you have questions on this problem and solution or if you have another question you would like to see me answer, just ask it.  Or if you have an entire topic you would like to see me write a lesson about, just let me know.

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Continuity

Continuity is a relatively simple concept, but problems that require proving it can be a little tricky. Essentially, a continuous function is one that you can draw all in one motion without picking up your pencil. This is one explanation of what it means for a function to be continuous that I like because it doesn’t take any mathematical definitions or proofs to understand. Any holes or gaps in a function’s graph would be a discontinuity and would mean that the function is not continuous.

The limit definition of continuity

By definition, a function $$f(x)$$ is continuous at $$x=a$$ if $$\lim_{x \to a} f(x) = f(a).$$

Let’s think about what this equation is saying. The left side of this equation is something that we’ve already dealt with, limits. It’s asking us to find out what $$y$$ value we close in on as we travel along our function and close in on $$x=a$$. Keep in mind, $$x$$ is a variable here which represents the input of our function, $$f(x)$$, and $$a$$ is a constant. This means that $$a$$ represents some specific number, which could be any number.

The right side of the equation is simply asking us to plug that same $$a$$ value into $$f(x)$$ and take the $$y$$ value we get out.

In total, the above equation says that if we travel along a function and close in on a specific $$x$$ value, we should close in on the same $$y$$ value we would get if we simply plugged that $$x$$ value into the equation.

In other words, as we travel along a function toward a specific $$x$$ value, our $$y$$ value will also go toward the $$y$$ value of the function at that point. If that is the case, then the function is continuous at that specific point. If we can say that a function is continuous at every single possible value we could put in for $$a$$, then we can say that the function is continuous for all $$x$$. If this is true then we can draw our entire function in one motion without picking up our pencil!

Let’s do a couple examples.

Example 1

Remember back in the first lesson about limits, Limits – Intro, I said I would go back to discussing the importance of our limit in the first example giving us the same value as when we plugged $$x=2$$ into the equation?  I would like to go into that further.

The function we were considering was $$f(x) = x^2$$ and we were finding

$$\lim_{x \to 2} x^2.$$

After looking at the graph of this function, shown in Figure 1.1, we saw that

$$\lim_{x \to 2} x^2 = 4.$$

I also pointed out that plugging in $$x=2$$ directly into the function also returns a $$y$$ value of $$4$$. In other words, we know $$f(2) = 4.$$ Therefore, we know

$$\lim_{x \to 2} x^2 = 4 = f(2)$$

$$\lim_{x \to 2} x^2 = f(2).$$

Notice this is exactly like the definition of what it means for a function to be continuous at a point. If you replace $$a$$ with $$2$$, replace $$f(x)$$ with $$x ^2$$, and replace $$f(a)$$ with $$4$$, we have just shown that $$y=x^2$$ is a function that is continuous at $$x=2.$$

Example 2

The next example I want to discuss also goes back to a function we have already looked at. Referring back to Limits – Intro we will consider $$f(x)$$ shown in Figure 1.2.

The question we will answer here is whether this function is continuous at $$x=1$$ or not. What do you think?

There are a few different ways we can answer this question. The simplest would be to simply look at the graph of the function and think about whether we can draw that function at and around $$x=1$$ without picking up our pencil. As you can see, there is clearly a hole at $$x=1$$ where we would need to pick up our pencil, and add a single point at $$(1, \ 4)$$.

As a result, it is probably safe to say that this function is not continuous at $$x=1$$. However, we want to be able to show this using the actual definition of what it means for a function to be continuous.

Remember, for this function, which we are calling $$f(x)$$ in this case, we need to be able to show that $$\lim_{x \to 1} f(x) = f(1).$$

If we can show this equation to be true, then $$f(x)$$ is continuous at $$x=1$$, and if it’s not true then the function is not continuous at $$x=1$$. Luckily, back when I first used this function as an example in the Limits – Intro lesson, we found that $$\lim_{x \to 1} f(x) = 2.$$ Therefore, we just need to find out if $$f(1)$$ is also $$2$$ and we can prove that this function is continuous or not continuous.

Looking at the graph again, we see that this function has a hole at $$x=1$$ and includes the point $$(1, \ 4)$$. In other words, if we plug $$1$$ into $$f(x)$$ as our $$x$$ value, we get a $$y$$ value of $$4$$ out. This is the same as saying $$f(1) = 4$$.

Now, at this point we have figured out $$\lim_{x \to 1} f(x) = 2 \ and$$ $$f(1) = 4.$$

Therefore, $$\lim_{x \to 1} f(x) \neq f(1)$$ and we can say that $$f(x)$$ is not continuous at $$x=1$$.

More Examples

Find the values of $$a$$ and $$b$$ that make $$f$$ continuous everywhere.

$$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

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There are also several other lessons and problems on my limits page.  It would be a good idea to get some practice with limits.  A lot of other more complex topics in calculus are based around limits so they are important to understand.  If you can’t find the topic you want to read about just let me know by emailing me at jakesmathlessons@gmail.com and I’ll do my best to answer your question!

Squeeze Theorem

The Squeeze Theorem is a useful tool for finding complex limits by comparing the limit to two much simpler limits. Squeeze Theorem tells us that if we know these three things:

$$1. \ \ \ g(x) \leq f(x) \leq h(x)$$

$$2. \ \ \ \lim_{x \to a} g(x) = L$$

$$3. \ \ \ \lim_{x \to a} h(x) = L$$

Then we also know that

$$\lim_{x \to a} f(x) = L$$

Keep in mind, requirement number 1 above only needs to be true around $$x=a$$. It does not need to be true at $$x=a$$ and it also does not need to be true when we get far away from $$x=a$$. This is because limits only care what is happening to a function when we get infinitely close to a given $$x$$ value, not at that $$x$$ value and certainly not far away from that $$x$$ value.

Example 1

You will typically see this used when you are finding the limit of a function which is made up of two functions being multiplied together and one of them is a trig function. For example, consider

$$\lim_{x \to 0}x ^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg).$$

At first glance, this limit looks terribly complicated. Especially because the function $$f(x) = x ^4 sin \Big( \frac{1}{\sqrt{x}} \Big)$$ doesn’t even exist at $$x=0$$ (plugging in $$x=0$$ directly causes you to divide by $$0$$, which can’t be done). However, if we use Squeeze Theorem, we can find this limit by finding two other limits that are much easier to find.

Squeeze Theorem says that we will need to find one function that is greater than or equal to $$f(x)$$ around $$x=0$$ (because we are finding the limit as $$x \to 0$$) and another function that is less than or equal to $$f(x)$$ around $$x=0$$. It is not always obvious which functions to use for comparison, but any function that is made up of $$sine$$ or $$cosine$$ of something multiplied by something else will usually follow the same pattern, so this is a good thing to try first.

What’s the pattern?

Consider the function $$y = sin(x)$$. No matter what you plug in for $$x$$ you will always get a $$y$$ value between $$-1$$ and $$1$$. This will be true regardless of what you are taking the $$sine$$ of. Even if you were to take $$sin\Big( \frac{1}{\sqrt{x}} \Big)$$. No matter what you put in for $$x$$ this will give you something between $$-1$$ and $$1$$, except if you try to put in $$x=0$$ (remember you can’t divide by $$0$$).

But remember, with limits it doesn’t matter what happens at the $$x$$ value we are approaching, only what is happening around that point. Therefore, it doesn’t matter that this function isn’t defined at $$x=0$$ because it is defined for all $$x$$ values near $$x=0$$.

The point I’m trying to make is that $$sine$$ of anything can be bound between $$-1$$ and $$1$$, so we can say:

$$-1 \leq sin(x) \leq 1, \ and$$

$$-1 \leq sin \bigg( \frac{1}{\sqrt{x}} \bigg) \leq 1$$

How does this relate to our example?

Now, the trick that makes this whole thing work is the fact that multiplying both sides of an inequality by a positive number maintains the inequality. Therefore, we can also multiply both sides of the inequality by a function that always outputs a positive number (or $$0$$).

For example, we can multiply both sides by $$x^4$$, or all three “sides” in this case. We can do this because any number raised to an even power will always be positive (or $$0$$). This gives us

$$x^4 \cdot \Bigg[ -1 \leq sin \bigg( \frac{1}{\sqrt{x}} \bigg) \leq 1 \Bigg]$$

$$x^4 \cdot (-1) \ \leq \ x^4 \cdot \Bigg(sin \bigg( \frac{1}{\sqrt{x}} \bigg) \Bigg) \ \leq \ x^4 \cdot (1)$$

$$-x^4 \ \leq \ x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) \ \leq \ x^4$$

Now, by using the Squeeze Theorem, since $$x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg)$$ is trapped between $$-x^4$$ and $$x^4$$ near $$0$$, if we find that if

$$\lim_{x \to 0} -x^4 = \lim_{x \to 0} x^4$$

then that will also be the answer for the limit we are looking for. Consider the graph of $$y=-x^4$$ and $$y=x^4$$ around $$x=0$$ shown below which was graphed using Desmos.

You can see that for both of these functions, as we travel along each function from both sides and get closer and closer to $$x=0$$, we get closer and closer to a $$y$$ value of $$0$$. This tells us that

$$\lim_{x \to 0} -x ^4 = 0, \ and$$

$$\lim_{x \to 0} x ^4 = 0$$

Putting it all together

These two facts in combination with the inequality we showed earlier: $$-x^4 \ \leq \ x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) \ \leq \ x^4,$$ tells us that

$$\lim_{x \to 0}x ^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) = 0.$$

Note: Compare these last 4 statements to the 4 statements at the beginning of this lesson.  The inequality is exactly the first thing I said we were looking to be able to show and the two limits are exactly like part 2 and 3.  So it’s no surprise that these three pieces combine to prove the limit we were trying to find.

Example 2

$$\mathbf{\lim\limits_{x \to \infty} \frac{sin(x)}{x}}$$ | Solution

The Squeeze Theorem is an important application of limits in calculus, but I have written several other lessons and problem solutions about limits!  You should go check out my limits page to see what else I’ve done.  If I haven’t answered any questions you have on that page, let me know by sending me an email at jakesmathlessons@gmail.com and I’ll do my best to address your questions.

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