Evaluate $\lim_{x \to 0+} (\sin x)^x$ I have tried to evaluate this by :--

$$(\sin x)^x = e^{(x \cdot \log (\sin x))} = (e^x)^{\log (\sin x)}$$

$$\lim_{x \to 0+} (e^x)^{\log (\sin x)} = \lim_{ x \to 0+} e^{\log (\sin x)}(e^x-1)$$

But after this I couldn't proceed.
Let $L=\lim\limits_{x\to0^+}(\sin x)^x$. Then $\ln L=\lim\limits_{x\to0^+}x\ln(\sin x)=\lim\limits_{x\to0^+}\frac{\ln(\sin x)}{1/x}$. This puts the limit in a form to use L'Hopital's rule:
$$\lim\limits_{x\to0^+}\frac{\ln(\sin x)}{1/x}=\lim\limits_{x\to0^+}\frac{\cos x/\sin x}{-1/x^2}=\lim\limits_{x\to0^+}-\frac{x^2\cos x}{\sin x}=-\lim\limits_{x\to0^+}\frac{x}{\sin x}\cdot\lim\limits_{x\to0^+}x\cos x$$
Using the fact that $\lim\limits_{x\to0^+}\frac{x}{\sin x}=1$ (this can be verified using the squeeze theorem), we get that $$-\lim\limits_{x\to0^+}\frac{x}{\sin x}\cdot\lim\limits_{x\to0^+}x\cos x=-1\cdot0\cdot1=0\implies \ln L=0$$
Now exponentiating:
$$\ln L=0\implies L=1$$
Therefore, $\lim\limits_{x\to0^+}(\sin x)^x=1$.
I like binomial expansions, so why not try one?

$$
\begin{align}
\lim_{x\rightarrow 0} (\sin(x))^x &amp; = \lim_{x\rightarrow 0}(\tfrac{e^{ix}-e^{-ix}}{2i})^x \\
&amp; = \lim_{x\rightarrow 0}\tfrac{1}{(2i)^x}\sum_{k=0}^{x} {x\choose k} e^{ix(x-k)}(-e)^{-ix(k)} \\
&amp; = \sum_{k=0}^{0} {0\choose k} e^{i0(0-k)}(-e)^{-i0(k)}\\
&amp; = {0\choose 0} e^{0}(-1)^0e^{0}\\
&amp; = 1
\end{align}
$$