Question

Solve : $\underset{x\to 0}{lim}\frac{e^x+e^{-x}-2}{x\tan x}$

Answer 1

$\underset{x\to 0}{lim}\frac{e^x+e^{-x}-2}{x\tan x}$

The given limit is of $\frac{0}{0}$ form.

So, apply L'Hospital Rule , i.e. differentiate Numerator and Denominator w.r.t. $x$

$=\underset{x\to 0}{lim}\frac{e^x-e^{-x}}{x{\sec }^{2}x+\tan x}$

Which is again $\frac{0}{0}$ form, so apply the rule again,

$=\underset{x\to 0}{lim}\frac{e^x+e^{-x}}{2x{\sec }^{2}x\tan x+{\sec }^{2}x+{\sec }^{2}x}$

$=\frac{1+1}{0+1+1}=1$