Question

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

Answer 1

Consider the given expression.

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

On differentiating with respect to x, we get

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

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

${lim}_{x⟶0}{f}^{\prime }\left(x\right)={lim}_{x⟶0}\frac{2\sin x-2\sin x\cos x}{x^3}$

${lim}_{x⟶0}{f}^{\prime }\left(x\right)={lim}_{x⟶0}\frac{2\sin x(1-\cos x)}{x^3}$

${lim}_{x⟶0}{f}^{\prime }\left(x\right)={lim}_{x⟶0}\frac{2\sin x}{x}\frac{(1-\cos x)}{x^2}$

${lim}_{x⟶0}{f}^{\prime }\left(x\right)={lim}_{x⟶0}\frac{2\sin x}{x}\frac{\left[2{\sin }^2\frac{x}{2}\right]}{x^2}$

${lim}_{x⟶0}{f}^{\prime }\left(x\right)={lim}_{x⟶0}\frac{\sin x}{x}\frac{{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}$

We know that,

${lim}_{x⟶0}\frac{\sin x}{x}=1$

Therefore,

${lim}_{x⟶0}{f}^{\prime }\left(x\right)=1\times 1$

${lim}_{x⟶0}{f}^{\prime }\left(x\right)=1$

Hence, this is the correct answer.