夹逼定理

Squeeze Theorem

This lesson will cover an theorem used to evaluate limits called the Squeeze Theorem. Essentially, we will use the squeeze theorem to determine how to evaluate $$\lim_{x \to 0} \frac{\sin x}{x}.$$

The basic idea behind the squeeze theorem is the following:

If $\forall x f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x)$, then it follows that $\lim_{x \to a} g(x) = L$.

Allow me to explain. $f(x)$ and $h(x)$ form the upper and lower bounds for $g(x)$, as in $g(x)$ can never be greater than $h(x)$ and can never be less than $f(x)$. So it stands to reason that if $f(x)$ and $h(x)$ approach the same value, $g(x)$ must approach that value too!

It's like the upper and lower bounds "squeeze" $g(x)$ and force it to have the same value. Here is a good illustration of this idea:

The functions shown in the diagram are $f(x) = -x^2$, $h(x) = x^2$, and $g(x) = 0$.

Although it is really obvious what the limit of $g(x)$ as $x$ approaches zero is $0$ since we can just substitute, we can also apply the Squeeze Theorem to compute the limit. We know that for all x, $f(x) \leq g(x) \leq h(x)$ and additionally $\lim_{x \to 0} x^2 = 0 = \lim_{x \to 0} -x^2$ so by the Squeeze Theorem we can say $\lim_{x \to 0} 0 = 0$.

Basically $f(x)$ and $h(x)$ squeeze $g(x)$ and force it to be the same value when $f(x) = h(x)$.

Limit of sin(x)/x

That last example of the squeeze theorem was trivial, but now we will use it to derive one of the most critical limits in all of calculus:

$\lim_{x \to 0} \frac{\sin x}{x}$

Ok. So to apply the Squeeze Theorem, we need to come up with the $f(x)$ and $h(x)$ to squeeze our $g(x)$.

Firstly, the limits of the sin function are $-1 \leq \sin(x) \leq 1$

Within these bounds we know that $\cos(x) \leq \frac{\sin x}{x}$

Moreover, from our definition of the sin function, we know $\sin(x) \leq 1$

Take a look the graph of these three functions and you will start to see the squeeze:

Clearly you can see that the blue line is being squeezed by the other two as x approaches zero. Now, we apply the squeeze theorem!

We know $\lim_{x \to 0} \cos(x) = 1$ by substitution since $\cos(0) = 1$!

Also $\lim_{x \to 0} 1 = 1$ since the function is constant.

Therefore, since $\cos(x) \leq \frac{\sin(x)}{x} \leq 1$, we know that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ by the Squeeze Theorem!

So that's the Squeeze Theorem!

Ok, you may be thinking that squeeze theorem is the name of the game, especially if you've looked at any of the other explanations for squeeze theorem on expii. However, there are other techniques that may serve the purpose of squeeze theorem; one such technique is l'Hopital's rule. Now, l'Hopital's rule is used on limits that are in a indeterminate form. So if you are solving for a limit that is a quotient and end up getting $0/0$ or positive or negative $\infty/\infty$, you are getting the wrong answer.

L'Hopital's rule states that, if the limit becomes one of these indeterminate forms: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

So using this knowledge, let's turn to a common squeeze theorem example to demonstrate how convenient l'Hopital's rule is.

Here we have: $\lim_{x \to 0} \frac{\sin x}{x}.$

Evaluating this normally gives us $\frac{0}{0}$. So, using l'Hopital's rule, we can take the derivatives of both the top and bottom to make the problem easier.

If $f(x)= \sin(x)$, then $f'(x)= \cos(x)$, and if $g(x)=x$ then $g'(x)=1$.

Now, putting these values into l'Hopital's rule, we get $\lim_{x \to 0} \frac{\cos x}{1}$, and if we then go ahead and plug 0 into cos (x), we get 1.

So that would mean $\lim_{x \to 0} \frac{\cos x}{1} = \lim_{x \to 0} \frac{1}{1}$, which means $\lim_{x \to 0} \frac{\cos x}{1} = 1$.

As you can see, this rule is very helpful and can be utilized in place of squeeze theorem to solve many limits.

(Videos) The Squeeze Theorem for Limits

by PatrickJMT

Here's a video by patrickJMT showing you what the squeeze theorem is and how to solve problems using it.

Say we have a function $f(x)$ where at the $x$-coordinate $a$ we have the $y$-coordinate $L$.

Now suppose we have another function, $g(x)$, that intersects $f(x)$ at the $x$-coordinate $a$.

Right now we have a function $f(x)$ that's greater than or equal to a function $g(x)$ and we're going to squeeze one more function in-between them. $$g(x)\leq\ \ \ \leq f(x)$$ Also notice that the limit of $g(x)$ as $x$ approaches $a$ is $L$ and that's the same as $f(x)$. $$\lim_{x\rightarrow a}g(x)=L=\lim_{x\rightarrow a}f(x)$$

Now suppose we have a third function that's trapped between the two.

So what we have is $$g(x)\leq h(x)\leq f(x)$$ So what can we say about the limit of this new function? Well since it's squeezed between $f(x)$ and $g(x)$ we can say $$\lim_{x\rightarrow a}h(x)=L$$ The idea is that if we have two outer functions that have a defined limit at a point, then any function squeezed between those two will also have the same limit at that point.

The first example he looks at is $$\text{If }\ 3x\leq f(x)\leq x^3+2\\ \text{for }\ 0\leq x\leq2\text{, evaluate:}\\ \lim_{x\rightarrow1}f(x)$$

So we don't know what the limit is for the function $f(x)$, but we can find the limit of the outer functions at this point. $$\begin{align} &\lim_{x\rightarrow1}3x=3\\ &\lim_{x\rightarrow1}x^3+2=3 \end{align}$$ And since $$\lim_{x\rightarrow1}3x\leq\lim_{x\rightarrow1}f(x)\leq\lim_{x\rightarrow1}x^3+2\\ 3\leq\lim_{x\rightarrow1}f(x)\leq3$$ So therefore, $$\lim_{x\rightarrow1}f(x)=3$$

The next example is the typical kind you might find on homework or tests. The problem states to find: $$\lim_{x\rightarrow0}x^2\cdot\cos\frac{1}{x^2}$$ Notice that we can't just plug in $x=0$ because we get $0\cdot\cos\frac10$ and $\frac10$ is undefined.

Well we know that cosine is between one and negative one. $$-1\leq\cos\frac1{x^2}\leq1$$ No matter what we plug in here we get out a value between these two numbers.

If we multiply the inside by $x^2$ we also have to multiply the outsides to keep it even, so this gives us $$\color{red}{x^2}\cdot-1\leq\color{red}{x^2}\cdot\cos\frac1{x^2}\leq\color{red}{x^2}\cdot x^2\\ -x^2\leq x^2\cdot\cos\frac1{x^2}\leq x^2$$ In this problem we had to come up with our outside functions on our own. But now that we have them we can find the limits of the outside. $$\lim_{x\rightarrow0}-x^2\leq\lim_{x\rightarrow0}x^2\cdot\cos\frac1{x^2}\leq\lim_{x\rightarrow0}x^2\\ 0\leq \lim_{x\rightarrow0}x^2\cdot\cos\frac1{x^2}\leq0$$ So therefore, $$\lim_{x\rightarrow0}x^2\cdot\cos\frac1{x^2}=0$$