# How do you use the Squeeze Theorem to show that sqrt (x) * e^(sin(pi/x))=0 as x approaches zero?

Oct 11, 2015

$- 1 \le \sin \left(\frac{\pi}{x}\right) \le 1$ for all $x \ne 0$.

So, ${e}^{-} 1 \le {e}^{\sin} \left(\frac{\pi}{x}\right) \le {e}^{1}$ for all $x \ne 0$.
(${e}^{u}$ is an increasing function)

For $x > 0$, $\sqrt{x}$ is defined and positive, so

$\frac{\sqrt{x}}{e} \le \sqrt{x} {e}^{\sin} \left(\frac{\pi}{x}\right) \le e \sqrt{x}$

${\lim}_{x \rightarrow {0}^{+}} \frac{\sqrt{x}}{e} = 0$ $\text{ }$ and $\text{ }$ ${\lim}_{x \rightarrow {0}^{+}} e \sqrt{x} = 0$.

Therefore, by the Squeeze Theorem,

${\lim}_{x \rightarrow {0}^{+}} \sqrt{x} {e}^{\sin} \left(\frac{\pi}{x}\right) = 0$