Question

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

• A. 0
• B. $\frac{1}{2}$
• C. 2
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{2}$

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

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

$=\underset{x\to 0}{lim}\frac{\tan x}{x}.\frac{2{\sin }^2\left(x/2\right)}{x^2}$

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