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Q. $ \displaystyle\lim_{x\to0} $ $ \frac{\cos (\sin \,x)-1 }{x^2} $ equals

AMUAMU 2019

Solution:

$\displaystyle\lim _{x \rightarrow 0} \frac{\cos (\sin x)-1}{x^{2}}$
Now, using L-Hopital's rule,
$=\displaystyle\lim _{x \rightarrow 0} \frac{-\sin (\sin x) \cdot \cos x}{2 x}$
Again using L-Hopital's rule
$=\displaystyle\lim _{x \rightarrow 0} \frac{-\left[-\sin (\sin x) \times \sin x+\cos ^{2} x\right.}{\times \cos (\sin x)]} $
$=-\frac{1}{2}$