Limit with L'Hospital with infinite indeterminate formats I'm trying to find the limit:
$$\large \lim_{x\to0}(\sin x)^x$$
Whst I did was apply L'Hospital Rule:
$$\large \text{let }y =(\sin x)^x\implies
\ln y=x\ln\sin x$$
$$\large 
\lim_{x\to0}\ln y
=
\lim_{x\to0} x\ln\sin x
=
\lim_{x\to0}\frac x{\frac1{\ln\sin x}}
=
\lim_{x\to0} \frac1{\frac {-\cos x}{(\ln\sin x)^2\sin x}}
=
\lim_{x\to0} (-\tan x )(\ln\sin x)^2
=
\lim_{x\to0} \frac{(\ln\sin x)^2}{\frac1{(-\tan x )}}
=
...
$$
Ultimately it keeps on going, please help me.

Whgen evaluating the limit of a product with l'Hopital's rule, I'd advise to put in the denominator the simplest factor. So:
$$\begin{align}
\lim_{x\to0} x\ln\sin x&=\lim_{x\to0}\frac{\ln\sin x}{1/x}\\
&=-\lim_{x\to0}x^2\cot x\\
&=\lim_{x\to0}\frac{x^2}{\tan x}\\
&=\lim_{x\to0}\frac{2x}{1+\tan^2x}\\
&=0
\end{align}$$
