Question

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

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

Answer 1

Answer: (C)

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