Evaluate $\lim\limits_{x \to \infty} \sin(\frac{1}{x})^x$

PRIMER:

In THIS ANSWER, I developed the pair inequalities that is introduced in elementary geometry

$$\bbox[5px,border:2px solid #C0A000]{\theta\cos(\theta) \le \sin(\theta) \le \theta} \tag 1$$

for $0\le \theta \le \pi/2$.

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Using $(1)$ with $\theta=1/x$, we have

$$\frac{\cos(1/x)}{x}\le \sin(1/x)\le \frac1x$$

for $\frac2\pi \le x$. Hence, we can write

$$\left(\cos(1/x)\right)^{x} \frac{1}{x^x}\le \left(\sin(1/x)\right)^x\le \frac1{x^x}$$

whence application of the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty} \left(\sin(1/x)\right)^x=0}$$

For $x>6/\pi,$ $0<\sin(1/x) <1/2.$ For such $x$ we have $0 < \sin(1/x)^x < (1/2)^x \to 0.$ Therefore the limit is $0$ by the squeeze theorem.