Calculate the following limit:
lim
I know the result must be \sqrt[3]{e} but I don't know how to get it. I've tried rewriting the limit as follows:
\lim_{x \rightarrow 0} e ^ {\ln {\left( \frac{\tan x}{x} \right) ^ \frac{1}{\sin^2 x}}} = \lim_{x \rightarrow 0} e ^ {\frac{1}{\sin^2 x} \ln {\left( \frac{\tan x}{x} \right)}}
From this point, I applied l'Hospital's rule but got 1 instead of \sqrt[3]{e}.
Thank you!
Answer
\lim_{x\to0}\frac{\log\frac{\tan x}x}{\sin^2x}\stackrel{l'H}=\lim_{x\to0}\frac{\frac x{\tan x}\frac{x\sec^2x-\tan x}{x^2}}{2\sin x\cos x}=\lim_{x\to0}\frac {\frac1{\sin x\cos x}-\frac1x}{2\sin x\cos x}=
=\lim_{x\to0}\frac{x-\sin x\cos x}{\underbrace{2x\sin^2x\cos^2x}_{=\frac x2\sin^22x}}\stackrel{l'H}=\lim_{x\to0}\frac{\overbrace{1-\cos^2 x+\sin^2x}^{2\sin^2x}}{\frac12\sin^22x+\underbrace{x\sin2x\cos2x}_{=x\sin4x}}\stackrel{l'H}=\lim_{x\to0}\frac{2\sin2x}{2\sin4x+4x\cos4x}=
\stackrel{l'H}=\lim_{x\to0}\frac{4\cos2x}{12\cos4x-16x\sin4x}=\frac4{12}=\frac13
and the limit is \;\;e^{1/3}\;
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