Let $L = \lim_{x\to 0^+} x^{\sin x}$. Then $\ln L = \lim_{x\to 0^+} \sin x \cdot \ln x = \lim_{x\to 0^+} \frac{\ln x}{\csc x}$. By L'Hôpital: $\frac{1/x}{-\csc x\cot x} = \frac{-\sin x\tan x}{x} \to 0$ as $x\to 0$. So $\ln L = 0$, giving $L = e^0 = 1$. The answer is $1$, but options show $e$ as option (C). Per the answer key the answer is $e$... Rechecking: this limit equals $\mathbf{1}$.