Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = {0^{\sin 0}} = {0^0} \cr
& {\text{This limit has the form }}{0^0}.{\text{ }} \cr
& {x^{\sin x}} = {e^{\sin x\ln x}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\sin x\ln x}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln x}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln x = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{\sin x}}}} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{\sin x}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{1}{x}}}{{ - \frac{{\cos x}}{{\sin x}}}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{x\cos x}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{x\cos x}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos x}}{{x\sin x + \cos x}} = \frac{0}{{0 + 1}} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\sin x\ln x}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln x}} = {e^0} = 1 \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = 1 \cr} $$