Applied Test Series

Question

 

What is the correct procedure to solve this limit ?

Comments

By taking log .

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Since, $\lim_{x\rightarrow 0} \frac{2(\sin^{-1}(x) -\tan^{-1}(x))}{x^3}=1$ [ By L’Hopital’s rule] and $\lim_{x\rightarrow 0}\frac{2}{x^2} =\infty$. So, it is $1^{\infty}$ form, So, answer will be $e^{\lim_{x\rightarrow 0}\left(\frac{2}{x^2}\left(\frac{2*( \sin^{-1} (x)- \tan^{-1} (x))}{x^3} -1\right)\right)} =\frac{1}{\sqrt{e}}$