Let $f(x)$ be the integral of $\frac{2\sin x - \sin 2x}{x^3}$ for $x
eq 0$. Find the value of $\li

Question

Let $f(x)$ be the integral of $\frac{2\sin x - \sin 2x}{x^3}$ for $x 
eq 0$. Find the value of $\lim_{x 	o 0} f'(x)$.

Professor PC Thomas Classes · Answer

By the Fundamental Theorem of Calculus, $f'(x) = \frac{2\sin x - \sin 2x}{x^3}$. We need to evaluate $\lim_{x 	o 0} \frac{2\sin x(1 - \cos x)}{x^3}$. This can be rewritten as $\lim_{x 	o 0} \left(2 \cdot \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2}ight)$. Using standard limits $\lim_{x	o 0} \frac{\sin x}{x} = 1$ and $\lim_{x	o 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$, the result is $2 	imes 1 	imes \frac{1}{2} = 1$.