Q. $\lim_{x \rightarrow 0}\frac{1 - \cos^{3}x}{x\sin{x\cos x}}$ is equal to
A) $\frac{2}{5}$
B) $\frac{3}{5}$
C) $\frac{3}{2}$
D) $\frac{3}{4}$
Explanation
(c)
$\lim_{x \rightarrow 0}\frac{1 - \cos^{3}x}{x\sin{x\cos x}}$
$= \lim_{x \rightarrow 0}\frac{\left( 1 - \cos x \right)(1 + \cos^{2}x + \cos{x)}}{x^{2}\cos x.\frac{\sin x}{x}}$
$= 3\lim_{x \rightarrow 0}\frac{1 - \cos x}{x^{2}}$
$= 3 \times \frac{1}{2}$
$= \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.