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Q. $\displaystyle\lim _{x \rightarrow 0} \frac{\cos (\tan x)-\cos x}{x^{4}}$ is equal to

Limits and Derivatives

Solution:

$\cos (\tan x)-\cos x=2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{\cos (\tan x)-\cos x}{x^{4}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right)}{x^{4}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right)}{x^{4}\left(\frac{x+\tan x}{2}\right)\left(\frac{x-\tan x}{2}\right)}\left(\frac{x^{2}-\tan ^{2} x}{4}\right)$
$=\frac{1}{2} \displaystyle\lim _{x \rightarrow 0} \frac{x^{2}-\tan ^{2} x}{x^{4}}$
$=\frac{1}{2} \displaystyle\lim _{x\rightarrow 0} \frac{x^{2}-\left(x+\frac{x^{3}}{3}+\frac{2}{15} x^{5}+\ldots\right)^{2}}{x^{4}}$
$=\frac{1}{2} \displaystyle\lim _{x \rightarrow 0} \frac{1}{x^{2}}\left(1-\left(1+\frac{x^{2}}{3}+\frac{2}{15} x^{4}+\ldots\right)^{2}\right)$
$=-\frac{1}{3}$