Question

The value of $\underset{y\to 0}{lim}\frac{(x+y)\sec (x+y)-x\sec x}{y}$ is equal to

• A. $\sec x(x\tan x+1)$
• B. $x\tan x+\sec x$
• C. $x\sec x+\tan x$
• D. None of these

Answer 1

The correct option is A $\sec x(x\tan x+1)$

Given $\underset{y\to 0}{lim}\frac{(x+y)\sec (x+y)-x\sec x}{y}$

$=\underset{y\to 0}{lim}\frac{x\sec (x+y)+y\sec (x+y)-x\sec x}{y}$

$=\underset{y\to 0}{lim}\left\{\frac{x\left\{\sec \right(x+y)-\sec x\}}{y}+\sec (x+y)\right\}$.......taking $x$ as common

$=\underset{y\to 0}{lim}\left[\frac{x}{y}\left\{\frac{\cos x-\cos (x+y)}{cos(x+y)cosx}\right\}\right]+\underset{y\to 0}{lim}\sec (x+y)$

$=\underset{y\to 0}{lim}\left[\frac{x2\sin (x+\frac{y}{2})\sin \left(\frac{y}{2}\right)}{y\cos (x+y)cosx}\right]+\sec x$

$=\underset{y\to 0}{lim}\left[\frac{x\sin (x+\frac{y}{2})}{\cos (x+y)\cos x}\times \frac{\sin \left(\frac{y}{2}\right)}{\frac{y}{2}}\right]+\sec x$

$=x\tan x\sec x+\sec x$

$=\sec x(x\tan x+1)$