Question

Evaluate
$L=\underset{x\to 0}{lim}\frac{1+x^2-\cos x}{x\tan x}$

Answer 1

$\underset{x\to 0}{lim}\frac{1+x^2-\cos x}{x\tan x}$ is of $\frac{0}{0}$ form.

Using L'Hospitals rule,we have

$=\underset{x\to 0}{lim}\frac{2x+\sin x}{x{\sec }^{2}x+\tan x}$ is of $\frac{0}{0}$ form.

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

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

$=\frac{2+1}{0+2}=\frac{3}{2}$