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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.

Did
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NotADeveloper
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2 Answers2

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$x^{-\sin x} = e^{-\sin x\ln x}=1-\sin x\ln x+o(x)$ 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 $x\rightarrow 0.$

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dezdichado
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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.

Paramanand Singh
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Rene Schipperus
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    "s it turns out l'Hopital's rule is not helpful, and actually there is a much easier answer" This is very often the case, actually one might even argue this is almost always the case. – Did Nov 19 '17 at 22:01
  • @did Yes, I agree, the method here of successively using standard limits to reduce to easier problems is quite successful. – Rene Schipperus Nov 19 '17 at 22:06
  • I have taken the liberty to fix a minor typo at the end. – Paramanand Singh Nov 20 '17 at 07:42