Question

Solve
$L=\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$

Answer 1

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$ is of $\frac{0}{0}$ form

Using L'Hospitals rule,we have

$=\underset{x\to 0}{lim}\frac{{\sec }^{2}x-\cos x}{3x^2}$ is of $\frac{0}{0}$ form

$=\underset{x\to 0}{lim}\frac{2{\sec }^{2}x\tan x+\sin x}{6x}$

$=\underset{x\to 0}{lim}\frac{2}{6}{\sec }^{2}x\frac{\tan x}{x}+\frac{1}{6}\frac{\sin x}{x}$

$=\frac{1}{3}\times 1\times 1+\frac{1}{6}\times 1$

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$