Find $\lim\limits_{x\to0}\frac 1x(x^{-\sin x}-(\sin x)^{-x})$ The question is to evaluate this limit:$$\lim_{x\to0}\frac{\big(\frac{1}{x}\big)^{\sin x}-\big(\frac{1}{\sin x}\big)^x}{x}$$
I tried using l'Hospital's rule, taking the logarithm, doing some manipulations using known limits, but without success.
 A: $x^{-\sin x} = e^{-\sin x\ln x}=1-\sin x\ln x+o(x)$ and $(\sin x) ^{-x} =e^{-x\ln\sin x} =1-x\ln\sin x+o(x).$ So your problem is equivalent to calculating: $$\lim\limits_{x\to 0}\dfrac{x\ln\sin x - \sin x\ln x}{x}.$$
But $\ln\sin x - \dfrac{\sin x \ln x}{x} = \ln\dfrac{\sin x}{x}+\ln x(1-\dfrac{\sin x}{x})=\ln\dfrac{\sin x}{x}+\ln x(\dfrac{x^2}{3!}-\dfrac{x^4}{5!}+...)\rightarrow 0,$ as $x\rightarrow 0.$
You would need to fill in the small details. 
A: I deleted my previous answer as there were (stupid) mistakes in it, but as it turns out l'Hopital's rule is not helpful, and actually there is a much easier answer. Write the expression as
$$\frac{e^{-(\ln x)(\sin x)}-e^{-x\ln \sin x}}{x}$$
$$=(\sin x)^{-x}\frac{e^{-(\ln x)(\sin x)+x\ln \sin x}-1}{x}$$
Now $(\sin x)^{x}\to 1$ is easily verified. The remainder can be written as, $$\frac{e^{f(x)}-1}{x}=\frac{e^{f(x)}-1}{f(x)}\frac{f(x)}{x}$$ where the first factor again limits to $1$. Thus we are reduced to evaluating
$$\ln(\sin x)-\frac{\sin x}{x}\ln x=\ln(\frac{\sin x}{x})-(x\ln x)(\frac{\sin x-x}{x^2})$$ and its easy to see that all elements limit to zero.
