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Q. $\displaystyle\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}$ is equal to

Limits and Derivatives

Solution:

$\displaystyle\lim _{y \rightarrow 0}\left\{\frac{x\{\sec (x+y)-\sec x\}}{y}+\sec (x+y)\right\}$
$=\displaystyle\lim _{y \rightarrow 0}\left[\frac{x}{y}\left\{\frac{\cos x-\cos (x+y)}{\cos (x+y) \cos x}\right\}\right]+\displaystyle\lim _{y \rightarrow 0} \sec (x+y)$
$=\displaystyle\lim _{y \rightarrow 0}\left[\frac{x 2 \sin \left(x+\frac{y}{2}\right) \sin \left(\frac{y}{2}\right)}{y \cos (x+y) \cos x}\right]+\sec x$
$=\displaystyle\lim _{y \rightarrow 0}\left[\frac{x \sin \left(x+\frac{y}{2}\right)}{\cos (x+y) \cos x} \times \frac{\sin \left(\frac{y}{2}\right)}{\frac{y}{2}}\right]+\sec x$
$=x \tan x \sec x+\sec x$
$=\sec x(x \tan x+1)$