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.

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.
