$\lim_{x \rightarrow 0^+}$ $(\sin x)^{\ln x}$ 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.

Using a well-known inequality, for $0 < x < 1$ we have
$$1-x < -\ln x < \frac{1- x}{x}. $$
Hence, for $0 < x < 1$ 
$$\ln x \ln \sin x= (-\ln x )(-\ln \sin x)> -\ln x (1-\sin x)>  - \ln x (1 - \sin 1)\to \infty$$
