How do you use the Squeeze Theorem to find lim tan(x)cos(sin(1/x)) as x approaches zero?

Sep 25, 2015

We need to consider right and left limits separately.

Explanation:

For all $x \ne 0$, we have $- 1 \le \cos \left(\sin \left(\frac{1}{x}\right)\right) \le 1$

Right Limit
For $0 < x < \frac{\pi}{2}$, we have $\tan x > 0$, so we can multiply the three parts of the inequality above by $\tan x$ without changing the inequalities.

For $0 < x < \frac{\pi}{2}$, we get

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

Since ${\lim}_{x \rightarrow {0}^{+}} \left(- \tan x\right) = 0 = {\lim}_{x \rightarrow {0}^{+}} \left(\tan x\right)$, we have (by the right hand squeeze theorem)
${\lim}_{x \rightarrow {0}^{+}} \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) = 0$

Left Limit
For $- \frac{\pi}{2} < x < 0$, we have $\tan x < 0$, so when we multiply the three parts of the inequality by $\tan x$ we must change the inequalities.

For $- \frac{\pi}{2} < x < 0$, we get

$- \tan x \ge \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) \ge \tan x$

Since ${\lim}_{x \rightarrow {0}^{-}} \left(- \tan x\right) = 0 = {\lim}_{x \rightarrow {0}^{-}} \left(\tan x\right)$, we have (by the right hand squeeze theorem)
${\lim}_{x \rightarrow {0}^{-}} \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) = 0$

Twi-sided Limit

Because both the right and left limits at $0$ are $0$, we conclude:

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