Question

If $f\left(x\right)=\int \frac{2\sin x-\sin 2x}{x^3}dx,x\ne 0$ then $\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $1/2$

Answer 1

The correct option is B $1$

$f^{{}^{\prime }}\left(x\right)=\frac{2\sin x-\sin 2x}{x^3}$

${lim}_{x\to 0}\frac{2\sin x-\sin 2x}{x^3}=\frac{2\sin x-2\sin x\cos x}{x^3}$

$=\frac{2\sin x(1-\cos x)}{x^3}$

$=\frac{4\sin x{\sin }^2\frac{x}{2}}{x^3}$

$=\frac{\sin x}{x}\times \frac{{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}=1$