Question

$\underset{x\to 0}{lim}\frac{\sin x-x+\frac{x^3}{6}}{x^5}$ is equal to

• A. $\frac{1}{120}$
• B. $\frac{1}{160}$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is A $\frac{1}{120}$

$\underset{x\to 0}{lim}\frac{\sin x-x+\frac{x^3}{6}}{x^5}$ ....... (1)

We know expansion of Sinx.
$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}+$ ........ (2)

Substituting (2) in (1) we get

$\underset{x\to 0}{lim}\frac{x^5}{5!.x^5}$

$=\frac{1}{5!}$

$=\frac{1}{120}$