If $f(x)$ is the integral function of the function $\frac{2\sin{x - \sin{2x}}}{x^{3}},\ x
eq 0$, then $\lim_{ ightarrow}{f^{'}(x)}$ is equal to

Question

If $f(x)$ is the integral function of the function $\frac{2\sin{x - \sin{2x}}}{x^{3}},\ x 
eq 0$, then $\lim_{ightarrow}{f^{'}(x)}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Since, $f(x)$ is the integral function of $\frac{2\sin{x - \sin{2x}}}{x^{3}}$, therefore by definition
$f^{'}(x) = \frac{2\sin{x - \sin{2x}}}{x^{3}}$
$\Rightarrow \ \ \ \ \lim_{x \rightarrow 0}\ f^{'}(x) = \lim_{x \rightarrow 0}\frac{2\sin{x - \sin{2x}}}{x^{3}}$
$= \lim_{x \rightarrow 0} = \frac{2\sin x}{x}.\frac{1 - \cos x}{x^{2}} = 1$

Q. If $f(x)$ is the integral function of the function $\frac{2\sin{x - \sin{2x}}}{x^{3}},\ x \neq 0$, then $\lim_{\rightarrow}{f^{'}(x)}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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