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. none of these

Answer 1

Answer: (A)

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