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 $0^0$.
 A: 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...
A: This limit does not exist. We can pick two sequences $a_n$ and $b_n$ such that $f(a_n)$ has different limit than $f(b_n)$.
\begin{align}
a_n &= 2\pi n + \frac{\pi}{2} \\
b_n &= 2\pi n - \frac{\pi}{2}
\end{align}
Then
$$
\left(1+\sin\left(2\pi n + \frac{\pi}{2}\right)\right)^{\frac1{2\pi n + \frac{\pi}{2}}} = 2^{\frac1{2\pi n + \frac{\pi}{2}}} \rightarrow 1
$$
and
$$
\left(1+\sin\left(2\pi n  - \frac{\pi}{2}\right)\right)^{\frac1{2\pi n  - \frac{\pi}{2}}} = 0^{\frac1{2\pi n - \frac{\pi}{2} }} =0 \rightarrow 0
$$
A: 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 $-1< sin x < 1$
0 < $1 + sinx$< 2 (Adding 1 on all sides)
$(log 0)/x < log [1+ sin x]/x < (log 2)/x$  (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: $\lim_{x \to \infty}$ $(1+ sin x)^{1/x}$ = $1$
