The Squeeze Theorem

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The Squeeze theorem allows us to exchange difficult functions for easy functions.

In mathematics, sometimes we can study complex functions by exchanging 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 $\displaystyle \lim _{x\to a} g(x) = L = \lim _{x\to a} h(x),$ then $\displaystyle \lim _{x\to a} f(x) = L$ .

I’m thinking of a function $f$ . I know that for all $x$ $0 \le f(x) \le x^2.$ What is $\displaystyle \lim _{x\to 0} f(x)$ ?

$f(x)$ $f(0)$ $0$ impossible to say

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}$

Does $\displaystyle \lim _{x\to 0}f(x)$ exist?

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

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

$-|\sqrt [5]{x}|\le f(x) \le |\sqrt [5]{x}|.$ Since $\lim _{x\to 0} -|\sqrt [5]{x}| = \answer [given]{0} = \lim _{x\to 0}|\sqrt [5]{x}|,$ we see by the Squeeze Theorem, Theorem, that $\displaystyle \lim _{x\to 0} f(x) = \answer [given]{0}$ .

The next example will involve right triangles and the unit circle. If you are unfamiliar with the unit circle and how it relates to right triangles or need a recap, check out this link or this link.

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

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. It will help to draw some diagrams to investigate this situation. Note that in the diagrams below, the red circle is the unit circle (a circle with radius 1).

Let’s call the triangle in the first diagram triangle $A$ , and the triangle in the final diagram triangle $B$ . From our diagrams above we can 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, $\displaystyle \lim _{\theta \to 0}\frac {\sin (\theta )}{\theta } = \answer [given]{1}$ .

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: $\displaystyle \lim _{x\to 0} \left (x^2\sin \left (\frac {1}{x}\right )\right )$

Let’s graph this function to see what’s going on. Use the + button to zoom in:

$\graph {f(x)=x^2*\sin (1/x)}$

The function $x^2\sin \left (\frac {1}{x}\right )$ has two factors:

$\overbrace {x^2}^{\text {goes to 0 as x goes to 0}} \cdot \underbrace {\sin \left (\frac {1}{x}\right )}_{\text {bounded between -1 and 1}}$

Hence we have $x^2 \cdot \answer [given]{-1} \le x^2\sin \left (\frac {1}{x}\right ) \le x^2 \cdot \answer [given]{1}$ and we see $\lim _{x\to 0} x^2 \cdot \answer [given]{-1} = \answer [given]{0} = \lim _{x\to 0} x^2 \cdot \answer [given]{1}$ and so by the Squeeze theorem, $\lim _{x\to 0}\left (x^2\sin \left (\frac {1}{x}\right )\right )=\answer [given]{0}.$