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.
 A: 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, $\lim\limits_{x\to0^+}(\sin x)^x=1$.
A: Apply L'Hopitals Rule twice.
$$ = \lim_{x \to 0^+} e^{\left(\frac{\log (\sin x)}{1/x}\right)}$$
$$ = \lim_{x \to 0^+} e^ {\left(\frac{-\cot x}{-1/x^2}\right)}$$
$$= \lim_{x \to 0^+} e^{\left(\frac{x^2}{\tan x}\right)}$$
$$= \lim_{x \to 0^+} e^{\left(\frac{2x}{(\sec x)^2}\right)}$$
$$= e^{0/1}=1$$
A: If you want a bit more than the limit itself, almost as you started, consider
$$y=\sin ^x(x)\implies \log(y)=x \log (\sin (x))$$ and use Taylor expansion
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\log (\sin (x))=\log (x)-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$ $$\log(y)=x \log (x)-\frac{x^3}{6}-\frac{x^5}{180}+O\left(x^7\right)$$
$$y=e^{\log(y)}=1+x \log (x)+\frac{1}{2} x^2 \log ^2(x)+O\left(x^3\right)$$
A: This is based on hint from user csch2 (via comments to question) and IMHO is a simpler approach to the problem.
The expression under limit can be written as $$x^x\left(\frac{\sin x} {x} \right) ^x$$ and the second factor tends to $1^0=1$ so that the desired limit is equal to the limit of $x^x$ as $x\to 0^+$. This is easily handled by taking logs to get the expression $x\log x$ which tends to $0$ so that $x^x\to 1$.
That $x\log x\to 0$ is a well known fact on par with other well known limits like $\dfrac{\log(1+x)}{x}\to 1$.
A: I like binomial expansions, so why not try one?
$$
\begin{align}
\lim_{x\rightarrow 0} (\sin(x))^x & = \lim_{x\rightarrow 0}(\tfrac{e^{ix}-e^{-ix}}{2i})^x \\
& = \lim_{x\rightarrow 0}\tfrac{1}{(2i)^x}\sum_{k=0}^{x} {x\choose k} e^{ix(x-k)}(-e)^{-ix(k)} \\
& = \sum_{k=0}^{0} {0\choose k} e^{i0(0-k)}(-e)^{-i0(k)}\\
& = {0\choose 0} e^{0}(-1)^0e^{0}\\
& = 1
\end{align}
$$
