Question

Using L' Hospital's rule, evaluate: $\underset{x\to 0}{lim}\frac{x-\sin x}{x^2\sin x}$.

Answer 1

Given, $\underset{x\to 0}{lim}\frac{x-\sin x}{x^2\sin x}=\underset{x\to 0}{lim}\frac{x-\sin x}{x^3}.\frac{x}{\sin x}$

$=\underset{x\to 0}{lim}\frac{x-\sin x}{x^3}.1$ $\left(\begin{array}{c}\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1\end{array}\right)$

Applying L' Hospital's Rule, we have
$=\underset{x\to 0}{lim}\frac{x-\sin x}{x^3}$ $...\frac{0}{0}$ form

$=\underset{x\to 0}{lim}\frac{1-\cos x}{3x^2}$ $...\frac{0}{0}$ form

$=\underset{x\to 0}{lim}\frac{\sin x}{6x}$ $...\frac{0}{0}$ form

$=\frac{1}{6}\underset{x\to 0}{lim}\frac{\sin x}{x}=\frac{1}{6}$.