Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

The Squeeze Theorem

In this section, we consider the evaluation of limits of the form

${\displaystyle \underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}}$

where

• the quotient of ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ is not continuous at $a$, and so Theorem 3.14 does not apply.
• the quotient cannot be simplified into a function to which Theorems 3.14, 3.17, 3.18, 3.20 or 3.21 can be applied.
• $\underset{x\to a}{\lim }f\left(x\right)=0$ and $\underset{x\to a}{\lim }g\left(x\right)=0\text{.}$

To analyse such limits, we may need to apply the “Squeeze Theorem” (see Theorem 3 on page 78 of the textbook).

Theorem 3.22. Squeeze Theorem.

If

$g\left(x\right)\le f\left(x\right)\le h\left(x\right)\text{for}x\to a\text{,}$

and

$\underset{x\to a}{\lim }g\left(x\right)=L=\underset{x\to a}{\lim }h\left(x\right)\text{,}$

then

$\underset{x\to a}{\lim }f\left(x\right)=L\text{.}$

This theorem holds if $x\to a$ is replaced by $x\to a^{+}$ or $x\to a^{-}\text{.}$

To apply this theorem, we must find two functions $g$ and $h$ to “squeeze” the function of the limit $f\text{.}$

For functions that involve sine and cosine, we use the fact that $-1\le \sin x\le 1$ and $-1\le \cos x\le 1$ for all real numbers $x\text{.}$

Example 3.55. The function ${\displaystyle \sqrt{x}\cos \left(\frac{1}{x}\right)}$ is continuous on the interval $\left(0,\infty \right)\text{.}$ [Why?]

To evaluate the limit ${\displaystyle \underset{x\to 0^{+}}{\lim }\sqrt{x}\cos \left(\frac{1}{x}\right)\text{,}}$ we apply the Squeeze Theorem.

Recall that $-1\le \cos \theta \le 1\text{.}$ We can multiply this inequality by $\sqrt{x}$ to get the function

${\displaystyle \sqrt{x}\cos \left(\frac{1}{x}\right)}$

in the middle:

${\displaystyle -\sqrt{x}\le \sqrt{x}\cos \left(\frac{1}{x}\right)\le \sqrt{x}\text{.}}$

Note that the inequality does not change, because $\sqrt{x}$ is positive for all $x>0\text{.}$

Since

$\underset{x\to 0^{+}}{\lim }\sqrt{x}=0=\underset{x\to 0^{+}}{\lim }-\sqrt{x}$

(by Theorem 3.14), we conclude that

${\displaystyle \underset{x\to 0^{+}}{\lim }\sqrt{x}\cos \left(\frac{1}{x}\right)=0}$

(by the Squeeze Theorem).

Note: The following example is very important; we will use it to find the derivative of trigonometric functions.

Example 3.56. An argument similar to the one used in Example 3.55 does not work when we attempt to evaluate the limit

${\displaystyle \underset{x\to 0}{\lim }\frac{\sin x}{x}\text{.}}$

[Try and see why.]

To find the two functions, we need to apply the Squeeze Theorem, we use the geometric argument given on page 146 of the textbook. You must read and understand this argument.

From the geometric argument, we see that

${\displaystyle \cos \theta <\frac{\sin \theta }{\theta }<1\text{for}0<\theta <\frac{\pi }{2}\text{.}}$

If ${\displaystyle -\frac{\pi }{2}<\theta <0}$, then ${\displaystyle 0<-\theta <\frac{\pi }{2}}$, and by the conclusion above, we have

$\cos \left(-\theta \right)<\frac{\sin \left(-\theta \right)}{-\theta }<1\text{.}$

Since $\cos \left(-\theta \right)=\cos \left(\theta \right)$ and $\sin \left(-\theta \right)=-\sin \left(\theta \right)$, we conclude that

${\displaystyle \cos \theta <\frac{\sin \theta }{\theta }<1\text{for}-\frac{\pi }{2}<\theta <0\text{.}}$

Therefore,

${\displaystyle \cos \theta <\frac{\sin \theta }{\theta }<1\text{for}-\frac{\pi }{2}<\theta <\frac{\pi }{2}\text{,}\text{except}\theta =0\text{.}}$

Theorem 3.14 tells us that $\underset{x\to 0}{\lim }\cos x=1=\underset{x\to 0}{\lim }1$, and by the Squeeze Theorem, we conclude that

${\displaystyle \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\text{.}}$