Find $\lim_{x\to0}|x|^{\sin x}$ How can I find that specific limit?
I've tried the following:
$f(x) = |x|^{\sin {x}} = e^{\sin x \cdot \ln(|x|)} = e^{\frac{\ln |x|}{\frac{1}{x}}}$
And use LLhopittal but I don't think it's correct since $\lim_{x\to0}\ln(|x|)=-\infty$ and $\lim_{x\to0}\frac{1}{x}=\infty$
 A: You do not need L'Hopital. Just rewrite
$$
f(x) = |x|^{\sin {x}} = e^{\sin x \cdot \ln(|x|)} =e^{\frac{\sin x}{x} \cdot x\ln(|x|)}\ ,
$$
and use the standard limits $$\sin(x)/x\to 1$$ and $$x\log |x|\to 0$$ as $x\to 0$, to conclude that your function tends to $e^0=1$.
A: The easiest way to get rid of the absolute value is using lateral limits:
$$\lim_{x\to 0^+} |x|^{\sin x}= \lim_{x\to 0^+} x^{\sin x}$$
now you can apply l'Hôpital normally.
A: Appears the indeterminate form $0^0.$ Using $\sin x\sim x$ as $x\to 0:$ $$\begin{aligned} \lambda=\lim_{x\to 0^+}\log |x|^{\sin x}=\lim_{x\to 0^+}\sin x\log x=\lim_{x\to 0^+}x\log x\\
=\lim_{x\to 0^+}\frac{\log x}{1/x}=\left\{\frac{-\infty}{+\infty}\right\}=\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+}\;(-x)=0.
\end{aligned}$$ $$\Rightarrow \lim_{x\to 0^+}|x|^{\sin x}=e^\lambda=1$$ 
On the other hand
$$\begin{aligned} \mu=\lim_{x\to 0^-}\log |x|^{\sin x}=\lim_{x\to 0^-}\sin x\log (-x)=\lim_{x\to 0^-}x\log (-x)\\
=\lim_{x\to 0^-}\frac{\log (-x)}{1/x}=\left\{\frac{-\infty}{+\infty}\right\}=\lim_{x\to 0^-}\frac{1/x}{-1/x^2}=\lim_{x\to 0^-}\;(-x)=0.
\end{aligned}$$ $$\Rightarrow \lim_{x\to 0^-}|x|^{\sin x}=e^\mu=1$$ So, $\displaystyle\lim_{x\to0}|x|^{\sin x}=1.$
A: $\lim_{x\to0}|x|^{\sin x}$
Getting rid of absolute value gives us 
$\lim_{x\to0+}(x^{\sin x})$
Remember this rule?
$$a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$$
$$\lim _{x\to \:0}\left(e^{\sin \left(x\right)\ln \left(x\right)}\right)$$
This is a better form I thik, that you can apply l'hoptial
