Limit of $\sqrt[x]{1+\sin x}$ with $x$ approaching to $0$ Could anyone explain to me how to calculate the limit $\lim_{x\to0} \sqrt[x]{1+\sin x}$? I am pretty sure that the answer is $e$, but how do I arrive at that? I tried to calculate one-sided limits playing with the Squeeze Theorem: in a neighbourhood $(0,\frac{\pi}{2})$ of $0$ one has $\sin x \leq x$, hence $\sqrt[x]{1+\sin x}\leq \sqrt[x]{1+x}$, and the right-hand side tends do $e$ when $x\to 0$. But how to evaluate the expression from the bottom?

Since $x^y$ is continuous near $x=e$ and $y=1$:
$$
\begin{align}
\lim_{x\to0}\left(1+\sin(x)\right)^{1/x}
&=\lim_{x\to0}\left((1+\sin(x))^{1/\sin(x)}\right)^{\lim\limits_{x\to0}\sin(x)/x}\\
&=e^1
\end{align}
$$

Another approach: take logs
$$
\begin{align}
\log\left(\lim_{x\to0}\left(1+\sin(x)\right)^{1/x}\right)
&=\lim_{x\to0}\frac{\sin(x)}{x}\frac1{\sin(x)}\log(1+\sin(x))\\
&=\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\frac1{\sin(x)}\log(1+\sin(x))\\
&=\lim_{x\to0}\frac{\sin(x)}{x}\log\left(\lim_{x\to0}(1+\sin(x))^{1/\sin(x)}\right)\\[6pt]
&=1\log(e)\\[12pt]
&=1
\end{align}
$$
