$f(x)$ is the integral of $\frac{2 \sin x-\sin 2 x}{x^{3}}, x
eq 0$ find $\lim _{x ightarrow 0}

Question

$f(x)$ is the integral of $\frac{2 \sin x-\sin 2 x}{x^{3}}, x 
eq 0$ find $\lim _{x ightarrow 0} f^{\prime}(x)$.

Professor PC Thomas Classes · Answer

Given, $f(x)=\int\left(\frac{2 \sin x-\sin 2 x}{x^{3}}ight) d x$\ $	herefore \quad f^{\prime}(x)=\frac{d}{d x}(f(x))=\frac{2 \sin x-\sin 2 x}{x^{3}}$\ $=\frac{2 \sin x-2 \sin x \cos x}{x^{3}}=\frac{2 \sin x}{x}\left(\frac{1-\cos x}{x^{2}}ight)$\ $\lim _{x ightarrow 0} f^{\prime}(x)=\lim _{x ightarrow 0} 2\left(\frac{\sin x}{x}ight)\left(\frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}ight)$\ Now $\lim _{x ightarrow 0} \frac{\sin ^{2} \frac{x}{2}}{x^{2}}=\frac{1}{4} \lim _{x ightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}ight)^{2}=\frac{1}{4}$\ $	herefore \quad \lim _{x ightarrow 0} f^{\prime}(x)=2 \cdot 1 \cdot \frac{2}{4}=1$\