Evaluate limit $\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}$ How to evaluate $$\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}?$$
Only idea I can think of is sandwich theorem, but then I get $0^0$.
Let $f(x)=(1+\sin{x})^{\frac{1}{x}}$. 

When $x= 2k \,\pi$ , ($k \in \mathbb{N}$),  $f(x)=1$

When $x= (2k -1/2)\,\pi $ , $f(x)=0$

Hence...
I went ahead and solved this using Sandwich Theorem but took logarithm first:

Given y = $(1+ sin x)^{1/x}$

Taking logarithm on both sides:

log y = $(1/x) log (1 + sinx)$

We know that $-1&lt; sin x &lt; 1$

0 &lt; $1 + sinx$&lt; 2 (Adding 1 on all sides)

$(log 0)/x &lt; log [1+ sin x]/x &lt; (log 2)/x$  (Taking Log on both sides and dividing by $x$)    

Now using LH Rule for computing the $\lim_{x \to \infty}$ $(log 0)/x$ :

We get the limit as 0 after differentiating the numerator and denominator. The numerator being a constant and denominator being differentitaed to 1, gives 0. 

Also, $\lim_{x \to \infty}$ $(log 2)/x$ = $0$

Hence $\lim_{x \to \infty}$ $log [1+ sin x]/x$ = $0$ by Sandwich Theorem.

Since $logy$ = $(1/x) log (1 + sinx)$,
from above we have;

$\lim_{x \to \infty}$ $logy$ = $0$

$implies$ $\lim_{x \to \infty}$ $y$ = $e^0$ = $1$

Answer: $\lim_{x \to \infty}$ $(1+ sin x)^{1/x}$ = $1$