# How do you use the Squeeze Theorem to find lim (1/x)cosx as x approaches infinity?

Nov 14, 2015

${\lim}_{x \rightarrow \infty} \left(\frac{1}{x}\right) \cos x = 0$

#### Explanation:

$- 1 \le \cos x \le 1$ for all $x$.

For $x > 0$, we have $\frac{1}{x} > 0$, so we can multiply the inequality by $1. x$ without reversing its direction.

$- \frac{1}{x} \le \left(\frac{1}{x}\right) \cos x \le \frac{1}{x}$ for $x > 0$.

${\lim}_{x \rightarrow \infty} \left(- \frac{1}{x}\right) = 0$ and ${\lim}_{x \rightarrow \infty} \left(\frac{1}{x}\right) = 0$.

Therefore, by the squeeze theorem (at infinity),

${\lim}_{x \rightarrow \infty} \left(\frac{1}{x}\right) \cos x = 0$