The Squeeze Theorem

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The Squeeze theorem allows us to compute the limit of a difficult function by ”squeezing” it between two easy functions.

In mathematics, sometimes we can study complex functions by relating them for simpler functions. The Squeeze Theorem tells us one situation where this is possible.

Squeeze Theorem Suppose that

$g(x) \le f(x) \le h(x)$

for all $x$ close to $a$ but not necessarily equal to $a$ . If

$\lim _{x\to a} g(x) = L = \lim _{x\to a} h(x),$

then $\lim _{x\to a} f(x) = L$ .

I’m thinking of a function $f$ . I know that for all $x$

$0 \le f(x) \le x^2.$

What is $\lim _{x\to 0} f(x)$ ?

Consider the function

$f(x) = \begin{cases} \sqrt [5]{x}\sin \left (\frac {1}{x}\right ) & \text {if $x \ne 0$,}\\ 0 & \text {if $x = 0$,} \end{cases}$

Is this function continuous at $x=0$ ?

We must show that $\lim _{x\to 0} f(x) = \answer [given]{0}$ . First, let’s assume that $x\ge 0$ and small. Since

$-1 \le \sin {\left (\frac {1}{x}\right )}\le 1$

by multiplying these inequalities by $\sqrt [5]{x} \ge 0$ , we obtain

$-\sqrt [5]{x} \le \sqrt [5]{x}\sin {\left (\frac {1}{x}\right )} \le \sqrt [5]{x}$

which can be written as

$-\sqrt [5]{x} \le f(x) \le \sqrt [5]{x},$

and, therefore, as

$-\Big |\sqrt [5]{x}\Big | \le f(x) \le \Big |\sqrt [5]{x}\Big |,$

Now, let’s assume that $x\le 0$ and small. Since

$-1\le \sin {\left (\frac {1}{x}\right )}\le 1$

by multiplying these inequalities by $\sqrt [5]{x} \le 0$ , we obtain

$\sqrt [5]{x} \le \sqrt [5]{x}\sin {\left (\frac {1}{x}\right )} \le -\sqrt [5]{x}$

which can be written as

$\sqrt [5]{x} \le f(x) \le -\sqrt [5]{x}.$

Recall that

$\Big |\sqrt [5]{x}\Big | = \begin{cases} -\sqrt [5]{x} & \text {if $x < 0$,}\\ \sqrt [5]{x} & \text {if $x \ge 0$.} \end{cases}$

Therefore for all small values of x

$-\Big |\sqrt [5]{x}\Big | \le f(x) \le \Big |\sqrt [5]{x}\Big |.$

Since

$\lim _{x\to 0} \left ( - \Big |\sqrt [5]{x}\Big | \right ) = \answer [given]{0} = \lim _{x\to 0}\Big |\sqrt [5]{x}\Big |,$

we apply the Squeeze Theorem and obtain that

$\lim _{x\to 0} f(x) = \answer [given]{0}$ . Hence $f(x)$ is continuous.

Here we see how the informal definition of continuity being that you can “draw it” without “lifting your pencil” differs from the formal definition.

Compute:

$\lim _{\theta \to 0} \frac {\sin (\theta )}{\theta }$

To compute this limit, use the Squeeze Theorem. First note that we only need to examine $\theta \in \left (\frac {-\pi }{2}, \frac {\pi }{2}\right )$ and for the present time, we’ll assume that $\theta$ is positive. Consider the diagrams below:

From our diagrams above we see that

$\text {Area of Triangle $A$} \le \text {Area of Sector} \le \text {Area of Triangle $B$}$

and computing these areas we find

$\frac {\cos (\theta )\sin (\theta )}{2} \le \frac {\theta }{2} \le \frac {\tan (\theta )}{2}.$

Multiplying through by $2$ , and recalling that $\tan (\theta ) = \frac {\sin (\theta )}{\cos (\theta )}$ we obtain

$\cos (\theta )\sin (\theta ) \le \theta \le \frac {\sin (\theta )}{\cos (\theta )}.$

Dividing through by $\sin (\theta )$ and taking the reciprocals (reversing the inequalities), we find

$\cos (\theta ) \le \frac {\sin (\theta )}{\theta } \le \frac {1}{\cos (\theta )}.$

Note, $\cos (-\theta ) = \cos (\theta )$ and $\frac {\sin (-\theta )}{-\theta } = \frac {\sin (\theta )}{\theta }$ , so these inequalities hold for all $\theta \in \left (\frac {-\pi }{2}, \frac {\pi }{2}\right )$ . Additionally, we know

$\lim _{\theta \to 0}\cos (\theta ) = \answer [given]{1} = \lim _{\theta \to 0}\frac {1}{\cos (\theta )},$

and so we conclude by the Squeeze Theorem, $\lim _{\theta \to 0}\frac {\sin (\theta )}{\theta } = \answer [given]{1}$ .

When solving a problem with the Squeeze Theorem, one must write a sort of mathematical poem. You have to tell your friendly reader exactly which functions you are using to “squeeze-out” your limit.

Compute:

$\lim _{x\to 0} \left (\sin (x) e^{\cos \left (\frac {1}{x^3}\right )}\right )$

Let’s graph this function to see what’s going on:

Remark: Since

$-1 \le \cos {\left (\frac {1}{x^3}\right )}\le 1,$

it follows that

$e^{-1} \le e^{\cos {\left (\frac {1}{x^3}\right )}}\le e^{1}.$

This remark is very important, since the function $\sin (x) e^{\cos \left (\frac {1}{x^3}\right )}$ has two factors:

Hence we have that when $0< x<\pi$

$0 \le \sin (x) e^{\cos \left (\frac {1}{x^3}\right )} \le \sin (x) \answer [given]{e}$

and we see

$\lim _{x\to 0^+}0 = \answer [given]{0} = \lim _{x\to 0^+} \sin (x) \answer [given]{e}$

and so by the Squeeze theorem,

$\lim _{x\to 0^+}\left (\sin (x)e^{\cos \left (\frac {1}{x^3}\right )}\right )=\answer [given]{0}.$

In a similar fashion, when $-\pi <x<0$ ,

$\sin (x) \answer [given]{e} \le \sin (x) e^{\cos \left (\frac {1}{x^3}\right )} \le 0$

and so

$\lim _{x\to 0^-}\sin (x) \answer [given]{e} =\answer [given]{0}=\lim _{x\to 0^-}0,$

and again by the Squeeze Theorem $\lim _{x\to 0^-}\left (\sin (x) e^{\cos \left (\frac {1}{x^3}\right )}\right )=0$ . Hence we see that

$\lim _{x\to 0}\left (\sin (x) e^{\cos \left (\frac {1}{x^3}\right )}\right )=\answer [given]{0}.$

2025-01-06 19:58:47

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