Question

If $f\left(x\right)=\int \frac{2\sin x-\sin 2x}{x^3}dx$ where $x\ne 0$ then $\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)$ has the value:

• A. $0$
• B. $1$
• C. $2$
• D. not defined

Answer 1

The correct option is B $1$

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

$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}dx$

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

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

$=\underset{x\to 0}{lim}\frac{2\sin x\times 2{\sin }^2\left(\frac{x}{2}\right)}{x^3}$

$=4\underset{x\to 0}{lim}\frac{\sin x}{x}\times \frac{{\sin }^2\left(\frac{x}{2}\right)}{x^2}$

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

$=4\times \frac{1}{4}\underset{x\to 0}{lim}\frac{\sin x}{x}\times {[\times \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}]}^2$

It is time to adjust limit of second function but don't change anything to first function.

If $x\to 0$, then $\frac{x}{2}$ to $\frac{0}{2}$, therefore, $\frac{x}{2}\to 0$.

$\therefore$ $x\to 0$ can be replaced by $\frac{x}{2}$ tends to $0$ to make the second eligible to use our limit formula.

$=\underset{x\to 0}{lim}\frac{\sin x}{x}\times \underset{\frac{x}{2}\to 0}{lim}{\left[\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right]}^2$

$=1\times (1{)}^2$

$=1$

$\therefore$ $f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}dx=1$