Question

$f\left(x\right)$ is the integral of $\frac{2\sin x-\sin 2x}{x^3},x\ne 0,$ find $\underset{x\to 0}{lim}{f}^{\prime }\left(x\right).$

Answer 1

$f^{\prime }\left(x\right)=\frac{2\sin x-\sin 2x}{x^3}$
$\therefore \underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}$
$=\underset{x\to 0}{Lt}2\frac{\sin x}{x}.\frac{(1-\cos x)}{x^2},[\because \sin 2x=2\sin x\cos x]$
$=\underset{x\to 0}{Lt}2\frac{\sin x}{x}.\frac{2{\sin }^2\left(x/2\right)}{x^2}$
$=\underset{x\to 0}{Lt}\frac{\sin x}{x}.{\left\{\frac{\sin \left(x/2\right)}{x/2}\right\}}^2=1.1=1.$