# How do you determine the limit of sin(x)/x as x approaches infinity?

Apr 8, 2018

${\lim}_{x \to \infty} \sin \frac{x}{x} = 0$

#### Explanation:

We'll use the Squeeze Theorem, which states that if we have some function $f \left(x\right) ,$ we define new functions $h \left(x\right) , g \left(x\right)$ such that

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

Then, we'll take ${\lim}_{x \to a} h \left(x\right) , {\lim}_{x \to a} g \left(x\right) .$

If these limits are equal, then ${\lim}_{x \to a} f \left(x\right) = {\lim}_{x \to a} h \left(x\right) = {\lim}_{x \to a} g \left(x\right)$

Or, all the limits are equal.

So, recall that

$- 1 \le \sin x \le 1$

Then,

$- \frac{1}{x} \le \sin \frac{x}{x} \le \frac{1}{x}$

Take ${\lim}_{x \to \infty} - \frac{1}{x} , {\lim}_{x \to \infty} \frac{1}{x} :$

${\lim}_{x \to \infty} - \frac{1}{x} = - \frac{1}{\infty} = 0$

${\lim}_{x \to \infty} \frac{1}{x} = \frac{1}{\infty} = 0$

So,

${\lim}_{x \to \infty} \sin \frac{x}{x} = 0$