# Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 12

$$\lim _{x \rightarrow 0^{+}} \sqrt{x} 4^{\cos \left(\frac{\pi}{x}\right)}=0$$

#### Work Step by Step

For every $x \neq 0$, we have $\cos \frac{\pi}{x} \in[-1,1]$, which means that for every $x \neq 0$ we have $4^{\cos \frac{\pi}{x}} \in\left[4^{-1}, 4^{1}\right]=\left[\frac{1}{4}, 4\right]$. Therefore, we can put $l(x)=\frac{1}{4} \sqrt{x}$ and $u(x)=4 \sqrt{x},$ which gives us $l(x) \leq f(x) \leq u(x)$ Then, by the Squeezing Theorem we have: $$\lim _{x \rightarrow 0^{+}} \sqrt{x} 4^{\cos \left(\frac{\pi}{x}\right)}=0$$

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