Question

The value of $\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin x\cos x}$ is?

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

Answer 1

Answer: (C)

$\frac{3}{2}$

The value of $\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin x\cos x}$ is
$=\underset{x\to 0}{lim}\frac{(1-\cos x)(1+\cos x+{\cos }^{2}x)}{x\sin x\cos x}$
$=\underset{x\to 0}{lim}\frac{2{\sin }^2\left(x/2\right)}{x\cdot 2\sin \left(x/2\right)\cos \left(x/2\right)}\times \frac{(1+\cos x+{\cos }^{2}x)}{\cos x}$
$=\underset{x\to 0}{lim}\left[\frac{\sin \left(x/2\right)}{2\left(x/2\right)}\times \frac{1+\cos x+{\cos }^{2}x}{\cos \left(x/2\right)\cos x}\right]$
$=\frac{1}{2}\times 3=\frac{3}{2}$