Is it true that $\,(1+\sin n)^{1/n}\to 1$? In order to show that $\lim_{n\to\infty}(1+\sin n)^{1/n}=1$, we need to show that $\,1+\sin n\,$ cannot not become too small. It suffices for example to show that
$$
1+\sin n\ge \frac{c}{n^k},
$$
for some $c,k>0$. This could be a consequence of showing that there exist $d,m>0$, such that
$$
\Big|\,\pi-\frac{p}{q}\,\Big|\ge \frac{d}{q^{m}}
$$
for every rational $p/q$ (cf. irrationality measure).
It would be very interesting if we could produce a more elementary proof.
 A: 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 
$\lim_{x\to\infty}f(x)=
\begin{cases}
0& x=\frac{3\pi}{2}+2\pi k\\
1&\text{otherwise}
\end{cases}$
