Squeeze Archives | Jake's Math Lessons

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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