Question

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

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

Answer 1

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

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

Expanding $\tan x$ and $\sin x$

$L={lim}_{x\to 0}\frac{(x+\frac{1}{3}{x}^3+.....)-(x-\frac{x^3}{3!}+.....)}{x^3}{lim}_{x\to 0}\frac{x+\frac{x^3}{3}+.........-x+\frac{x^3}{3!}......}{x^3}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}\therefore L=\frac{1}{2}$