Question

State the whether given equation is true or false

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}=\frac{1}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

Solution -

${lim}_{x\to 0}\frac{tanx-sinx}{x^3}=\frac{1}{2}$ prove.

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

Applying L Hospital Rule. 3 times.

${lim}_{x\to 0}\frac{se{c}^{2}x-cosx}{3x^2}$

${lim}_{x\to 0}\frac{2se{c}^{2}xtanx+sinx}{6x}$

${lim}_{x\to 0}\frac{4se{c}^{2}xta{n}^{2}x+2se{c}^{4}x+cosx}{6}$

$=\frac{2+1}{6}=\frac{1}{2}=RHS$

Hence proved.