Question

$\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. does not exist

Answer 1

The correct option is A $1$

$\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}$
$=\underset{x\to 0}{lim}\frac{2\sin x(1-\cos x)(1+\cos x)}{x^3(1+\cos x)}$
$=\underset{x\to 0}{lim}2\frac{{\sin }^{3}x}{x^3}\times \frac{1}{1+\cos x}$
$=2\times (1{)}^3\times \frac{1}{1+1}=1$