Answer
$1$
Work Step by Step
To find the limit
$$
\lim _{x \rightarrow 0^+}x^{\sin x}
$$
we have to find
$$\ln \lim _{x \rightarrow 0^+}x^{\sin x}=\lim _{x \rightarrow 0^+}\ln x^{\sin x}\\
=\lim _{x \rightarrow 0^+}\frac{\ln x}{\csc x}=\frac{\infty}{\infty}$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0^+}\frac{\ln x}{\csc x}=\lim _{x \rightarrow 0^+}\frac{1/x}{-\csc x\cot x}\\
=\lim _{x \rightarrow 0^+}\frac{-\sin^2 x}{x\cos x}=\frac{0}{0}
$$
Which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow 0^+}\frac{-\sin^2 x}{x\cos x}=\lim _{x \rightarrow 0^+}\frac{-2\sin x\cos x}{\cos x-x\sin x}=\frac{0}{1}=0.$$
Now, $$\ln \lim _{x \rightarrow 0^+}x^{\sin x}=0\Longrightarrow \lim _{x \rightarrow 0^+}x^{\sin x}=e^0=1. $$
