Question

If $f\left(x\right)$ is the integral of $\frac{2\sin x-\sin 2x}{x^3},x\ne 0$. Find ${lim}_{x\to 0}{f}^{{}^{\prime }}\left(x\right)$, where $f^{{}^{\prime }}\left(x\right)=\frac{df\left(x\right)}{dx}$

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{1}{3}$
• D. $2$

Answer 1

The correct option is B $1$

$f\left(x\right)=\frac{2\sin x-\sin 2x}{x^3}$ $x\ne 0$

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

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

$\Rightarrow \underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}2\left(\frac{\sin x}{x}\right)\left(\frac{2{\sin }^2\left(x/2\right)}{x^2}\right)$

$\Rightarrow 4\cdot 1\underset{x\to 0}{lim}\left(\frac{{\sin }^2\left(x/2\right)}{4\times (x/2{)}^2}\right)$

$=1$