Question

Evaluate : ${lim}_{x\to 0}\frac{x-\sin x}{x^3}$

Answer 1

We have,

${lim}_{x\to 0}\frac{x-\sin x}{x^3}$

Apply L-hospital rule,

${lim}_{x\to 0}\frac{1-\cos x}{3x^2}$

Apply L-hospital rule,

${lim}_{x\to 0}\frac{0+\sin x}{6x}$

$\frac{1}{6}{lim}_{x\to 0}\frac{\sin x}{x}$

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

$=\frac{1}{6}$

Hence, this is the answer.