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 $0^0$.

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
$$
\left(1+\sin\left(2\pi n  - \frac{\pi}{2}\right)\right)^{\frac1{2\pi n  - \frac{\pi}{2}}} = 0^{\frac1{2\pi n - \frac{\pi}{2} }} =0 \rightarrow 0
$$
