Question

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

• A. $\sec x(x\tan x+1)$
• B. $x\tan x+\sec x$
• C. $x\sec x+\tan x$
• D. $x\sin x$

Answer 1

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

$\underset{y\to 0}{lim}\{\frac{x\sec (x+y)-\sec x}{y}+\sec (x+y\left)\right\}$
$\underset{y\to 0}{lim}\left[\frac{x}{y}\left\{\frac{\cos x-\cos (x+y)}{\cos (x+y)\cos x}\right\}\right]+\underset{y\to 0}{lim}\sec (x+y)$
$\underset{y\to 0}{lim}\left[\frac{x2\sin (x+\frac{y}{2})\sin \left(\frac{y}{2}\right)}{y\cos (x+y)\cos x}\right]+\sec x$
$\underset{y\to 0}{lim}[\frac{x\sin (x+\frac{y}{2})}{\cos (x+y)\cos x}\times \frac{\sin \frac{y}{2}}{\frac{y}{2}}]+\sec x$
$=x\tan x\sec x+\sec x$
$=\sec x(x\tan x+1)$