$\lim_{x \rightarrow 0^+}$ $(\sin x)^{\ln x}$ My question is, what is $\lim_{x  \rightarrow 0^+}(\sin x)^{\ln x}$?
This limit is equal to
$$\lim_{x \rightarrow 0^+} e^{\ln(\sin x)\ln x} = e^{\lim_{x\to0^+} \ln(\sin x)\ln x}$$
But what is right hand side limit of $\ln(\sin x)\ln x$  when x goes to $0$ 
any suggestion? I cannot apply L'Hospital.

Outline:
First, when $x\to 0^+$, we have $\sin x > 0$ for $x$ sufficiently small and the quantity is well-defined. Now (the following will rely on Taylor approximations):
$$
\sin x = x + o(x^2)
$$
so
$$
\ln \sin x = \ln( x + o(x^2))  = \ln x + \ln (1+o(x)) = \ln x + o(x)
$$
and 
$$
e^{\ln( \sin x) \cdot \ln x} = e^{(\ln x)^2 + o(x\ln x)}= e^{(\ln x)^2 + o(1)}
$$
as $x\ln x \xrightarrow[x\to 0]{} 0$. To conclude, observing that $(\ln x)^2 \xrightarrow[x\to 0]{} \infty$ will give you the answer by continuity of $\exp$.
