Question

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

• A. $\frac{1}{2}$
• B. $-\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is B $-\frac{1}{6}$

$\underset{x\to 0}{lim}\frac{\sin x-x}{x^3}=\underset{x\to 0}{lim}\frac{\{x-\frac{x^3}{3!}+\frac{x^5}{5!}....\}-x}{x^3}$
$\underset{x\to 0}{lim}\{\frac{1}{3!}+\frac{x^2}{5!}....\}=-\frac{1}{3!}=-\frac{1}{6}$

Alternate Solution
$\underset{x\to 0}{lim}\frac{\sin x-x}{x^3}\to \frac{0}{0}\text{form}$
Apply L'hospital rule
$\underset{x\to 0}{lim}\frac{\cos x-1}{3x^2}\to \frac{0}{0}\text{form}$
Apply L'hospital rule
$\underset{x\to 0}{lim}\frac{-\sin x}{6x}\to \frac{0}{0}\text{form}$
$\underset{x\to 0}{lim}\frac{-\cos x}{6}$
$=-\frac{1}{6}$