Question

Evaluate : ${lim}_{y\to 0}\frac{(x+y)sec(x+y)-xsecx}{y}$

Answer 1

$\underset{y\to 0}{lt}\frac{(x+y)\sec (x+y)-xsecx}{y}$ ($÷$ form)

$=\underset{y\to 0}{lt}\frac{\sec (x+y)+(x+y)\sec (x+y)\tan (x+y)}{1}$ (Using $L^1$ Hospital's rule)

$=\underset{y\to 0}{lt}\sec (x+y)[1+(x+y)\tan (x+y)]$

$=secx(1+xtanx)$