Question

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

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

Answer 1

The correct option is B $\frac{1}{2}$

$=\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$
$=\underset{x\to 0}{lim}\frac{\sin x(1-\cos x)}{x^3\cos x}$
$=\underset{x\to 0}{lim}\frac{\sin x\left(2{\sin }^2\frac{x}{2}\right)}{x^3\cos x}$
$=\frac{2}{4}=\frac{1}{2}$