Limit involving $\left |x \right |$

$y=\lim_{x \rightarrow 0} x^{\sin{x}}$

$\ln{y}=\lim_{x \rightarrow 0} (\sin{x}) (\ln{x})=\lim_{x \rightarrow 0} \dfrac{\ln{x}}{\csc{x}}$

Using L'Hopital's Rule :

$\cdots=\lim_{x \rightarrow 0} \dfrac{\frac{1}{x}}{-\cot{x} \csc{x}}=-\lim_{x \rightarrow 0} \dfrac{\sin^2{x}}{x \cos{x}}=-\left (\lim_{x \rightarrow 0} \dfrac{1}{\cos{x}} \right) \left (\lim_{x \rightarrow 0} \dfrac{\sin^2{x}}{x} \right)=-\lim_{x \rightarrow 0} \dfrac{\sin^2{x}}{x} \\ \cdots=-\lim_{x \rightarrow 0} \dfrac{2\sin{x} \cos{x}}{1}=-\lim_{x \rightarrow 0} \sin{2x}=0$

$\ln{y}=0 \Rightarrow y=e^0=\boxed{1}$

The limit $\displaystyle L = \lim_{x \to 0} \ |x|^{\sin x}$ exists if $\displaystyle \lim_{x \to 0^+} |x|^{\sin x} = \lim_{x \to 0^-} |x|^{\sin x}$ . Now consider:

$\begin{aligned} L_+ & = \lim_{x\to 0^+} |x|^{\sin x} \\ & = \lim_{x\to 0^+} x^{\sin x} \\ & = \lim_{x\to 0^+} \color{#3D99F6}{e}^{\ln x \sin x} \quad \quad \small \color{#3D99F6}{\text{Expressing } e^x \text{ by a more convenient form }\exp(x)} \\ & = \lim_{x\to 0^+} \color{#3D99F6}{\exp }(\ln x \sin x) \quad \quad \small \color{#3D99F6}{\text{Dividing up and down by }\sin x} \\ & = \lim_{x\to 0^+} \exp \left(\frac{\ln x}{\csc x}\right) \quad \quad \small \color{#3D99F6}{\text{Since } \ln x \to -\infty \text{ and }\csc x \to \infty \text{ as }x \to 0^+ \text{, we can use the L'Hôpital's rule.}} \\ & = \lim_{x\to 0^+} \exp \left(\frac{\frac{1}{x}}{-\csc x \cot x}\right) \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down}} \\ & = \lim_{x\to 0^+} \exp \left(\frac{\color{#3D99F6}{\sin x}\sin x}{\color{#3D99F6}{x}\cos x}\right) \\ & = \exp \left(\frac{(\color{#3D99F6}{1})(0)}{1}\right) \\ & = e^0 = 1 \end{aligned}$

Now, let us look at:

$\begin{aligned} L_- & = \lim_{x\to 0^-} |x|^{\sin x} \\ & = \lim_{x\to 0^\color{#D61F06}{+}} |\color{#D61F06}{-}x|^{\color{#D61F06}{-}\sin x} \\ & = \lim_{x\to 0^\color{#D61F06}{+}} x^{\color{#D61F06}{-}\sin x} \\ & = \lim_{x\to 0^\color{#D61F06}{+}} \frac 1 {x^{\sin x}} \\ & = \frac1L_+ = 1 \end{aligned}$

Therefore, $L = L_+ = L_- = \boxed{1}$

Easier way is to plot a graph of the given limit. See at zero, the limit is one.