Is it true that $\,(1+\sin n)^{1/n}\to 1$?

Consider $f(x)=(1+\sin x)^{\frac{1}{x}}$ . The binomial expansion is valid for $|\sin x|<1$.

Case I

$x\neq\frac{\pi}{2}+2\pi k$ and $x\neq\frac{3\pi}{2}+2\pi k$, where $k\in\mathbb{Z}$. Hence we may expand binomially

$f(x)=1+\frac{\sin x}{x}+\frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}\sin^2x+\dots+\frac{\frac{1}{x}(\frac{1}{x}-1)\dots(\frac{1}{x}-k+1)}{k!}\sin^kx+\dots$

EDIT (after comments made by Andréas)

$f(x)=1+\sum_{k=-x}^\infty\frac{(\frac{1}{x})!}{(x+k)!(\frac{1}{x}-x-k)!}\sin^{x+k}x$

Consider the coefficient of $\sin^{x+k}x$.

$g(x)=\frac{(\frac{1}{x})!}{(x+k)!(\frac{1}{x}-x-k)!}=\frac{\frac{1}{x}(\frac{1}{x}-1)\dots(\frac{1}{x}-x-k+2)(\frac{1}{x}-x-k+1)}{(x+k)!}$

Now taking the limit as $x\to\infty$,

\begin{align*} \lim_{x\to\infty}g(x) & =\lim_{x\to\infty}\frac{1}{x}\lim_{x\to\infty}\frac{(-1)(-2)\dots(-x-k+2)(-x-k+1)}{(x+k)!}\\ & =\lim_{x\to\infty}\frac{(-1)^{x+k-1}}{x}\lim_{x\to\infty}\frac{(x+k-1)!}{(x+k)!}\\ & =\lim_{x\to\infty}\frac{(-1)^{x+k-1}}{x(x+k)}\\ & =0 \end{align*}

Every term in the sum tends to zero, leaving $\lim_{x\to\infty}f(x)=1$.

Case II

$x=\frac{\pi}{2}+2\pi k$

Now $f(x)=2^{\frac{1}{x}}$ and clearly $\lim_{x\to\infty}f(x)=1$ again.

Case III

$x=\frac{3\pi}{2}+2\pi k$

Now $f(x)=0^{\frac{1}{x}}$ and therefore $\lim_{x\to\infty}f(x)=0$.

We may conclude that

$\lim_{x\to\infty}f(x)= \begin{cases} 0& x=\frac{3\pi}{2}+2\pi k\\ 1&\text{otherwise} \end{cases}$