Question

If $f\left(x\right)$ is the integral of the function $\frac{2\sin x-\sin 2x}{x^3},x\ne 0$

then $\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)$ is equal to?

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$1$

$f\left(x\right)=\int \frac{2\sin x-\sin 2x}{x^3}$ (given)
We want,
${lim}_{x\to 0}{f}^{\prime }\left(x\right)={lim}_{x\to 0}\frac{2\sin x-\sin 2x}{x^3}$
Substituting we get indeterminate form,
So, applying L'Hospital's Rule,
$={lim}_{x\to 0}\frac{2\cos x-2\cos 2x}{3x^2}$
Substituting we get indeterminate form so, applying L'Hospital's Rule,
$={lim}_{x\to 0}\frac{-2\sin x+4\sin 2x}{6x}$
Substituting we get indeterminate form so, applying L'Hospital's Rule,
$={lim}_{x\to 0}\frac{-2\cos x+8\cos 2x}{6}$
$=\frac{6}{6}=1$
Hence, option 'B' is correct.