Computing $\lim_{x\to 0^+} x^{\sin(x)}$ I'm trying to compute
$$
\lim_{x\to 0^+} x^{\sin(x)}.
$$
I've been trying to get it into a form where I can apply L'Hopital's rule, but I haven't had any success. Namely, I haven't been able to massage it into a fraction form. Any advice?
 A: Hint Using the L'Hopital's rule:
$$\lim_{x\to0}\sin x\ln x=\lim_{x\to0}\frac{\ln x}{\frac1{\sin x}}=-\lim_{x\to0}\frac{\sin x}{x}\frac{\sin x}{\cos x}=0$$
A: We have $\lim_{x \to 0^+} \sin(x) \log(x) = 0$. Hence,
$$\lim_{x \to 0^+} x^{\sin(x)} = 1$$
A: Write
$$x^{\sin x}=e^{\ln x\sin x}$$
Then, apply l'Hopital's rule to 
$$\frac{\sin x}{1/\ln x}$$
Last, apply the continuity of exponential function.
A: $\mathbf{Result \; I}:$ If $x \in [0,\pi)$, then 
$$ x^{\sin x} \leq \sin x(x - 1 ) + 1  $$
To see why this is true, notice that $x \in [0, \pi) $ implies that $\sin x \geq 0 $, hence invoking the Young's Inequality and the fact that $\sin x + (1 - \sin x) = 1$, we have
$$ x^{\sin x} = x^{\sin x}\cdot1^{1- \sin x} \leq \sin x \cdot x + 1 - \sin x = \sin x(x -1 ) + 1 $$
$\mathbf{Result \; II}$: $$x^{\sin x} \geq 1 - 10 x $$
You should try to show this inequality: You can find the min of $f(x) = x^{ \sin x} + 10 x $.
Now, 
$$ \sin x(x -1) + 1 \geq x^{\sin x} \geq 1 - 10 x $$
Hence, an application of of the squeeze rule gives that 
$$ \lim_{x \to 0^+} x^{\sin x} = 1 $$
