Problem getting to and understanding the answer of $\lim \limits_{x \to 0^+} \frac{(\sin{x})^x}{x}$ $$\lim \limits_{x \to 0^+} \frac{(\sin{x})^x}{x}$$
Somehow right limit ($0^+$) is $+\infty$ but left ($0^-$) one is undefined. I wrote limit like $ \lim \limits_{x \to 0^+} \frac{\sin{x}}{x} * \lim \limits_{x \to 0^+} (\sin{x})^{x-1}$ and the first one is 1 so it leaves us with just $\lim \limits_{x \to 0^+} (\sin{x})^{x-1}$. So if we put the values of the limit in: $\sin{x}$ when $x$ is converging to $0$ is $0$ and we get $0^{0-1}$ or $\frac{1}{0}$. So if the right limit is $\frac{1}{0^+} = +\infty$ why isn't $\frac{1}{0^-} = -\infty$?
 A: If I have a real function of real variable how this:
$$\psi(x)=f(x)^{g(x)}$$
your domain is $$\text{dom}(\psi(x))=\{x\in\Bbb R \mid f(x)>0\,\wedge x \in \operatorname{dom}(g(x))\} \tag 1$$ Considering the $(1)$, the function
$$h(x)=\frac{(\sin{x})^x}{x}$$
have a domain, $D$ with $x\neq 0 \ \wedge \ \sin(x)>0$. If you solved this $0^-\notin D$. Hence:
$$\lim_{x \to 0^-} \frac{(\sin{x})^x}{x}=\nexists$$
After, being $0<\sin(x)\leq 1$, and $x\to 0^+\implies$
$$\lim_{x \to 0^+} \sin^{x}{(x)} = \lim_{x \to 0^+} e^{\ln (\sin^{x} x)}=e^{\lim_{x \to 0^+} \frac{\ln{\sin x}}{\frac{1}{x}}} \tag 2$$
where in the ratio you have an indeterminal form $(\infty/\infty)$. You use the l'Hopital's rule. After you will have an indeterminal form $(0/0)$. Using another times the l'Hopital's rule you can try that you will have
$$e^{- \lim_{x \to 0^+} \frac{2 x}{\tan^{2}{x} + 1}} = e^{- \lim_{x \to 0^+} 2 x \cos^{2}{x}}=1$$
Therefore
$$\lim_{x \to 0^+} \sin^{x}{x}=1$$ and
$$\lim_{x \to 0^+} \frac{(\sin{x})^x}{x}=+\infty$$
A: Hint: use that $\sin x\geq\frac{x}{2}$ for $x$ close to $0$.
