Question

$\underset{x\to 0}{\lim }\frac{1-{\cos }^{3}x}{x\sin x\cos x}$ is equal to

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $\frac{3}{2}$
• D. $\frac{3}{4}$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{1-{\cos }^{3}x}{x\sin x\cos x}$
$=\underset{x\to 0}{\lim }\frac{2\left(1-{\cos }^{3}x\right)}{x\sin 2x}$
[Using LHospital Rule]
$=\underset{x\to 0}{\lim }\frac{2\left[-3{\cos }^{2}x(-\sin x)\right]}{\sin 2x+x\cos 2x\cdot 2}$
$=\underset{x\to 0}{\lim }\frac{6{\cos }^{2}x\sin x}{\sin 2x+2x\cos 2x}$
$=\underset{x\to 0}{\lim }\frac{6\left[-2\cos x{\sin }^{2}x+{\cos }^{3}x\right]}{2\cos 2x+2[-x\sin 2x\cdot 2+\cos 2x]}$
$=\underset{x\to 0}{\lim }\frac{6\left[-2\cos x{\sin }^{2}x+{\cos }^{3}x\right]}{2\cos 2x-4x\sin 2x+2\cos 2x}$
$=\frac{6}{2+2}$
$=\frac{3}{3}$