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

The correct option is C $\frac{3}{2}$

$\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin x\cos x}=\underset{x\to 0}{lim}\frac{(1-\cos x)(1+{\cos }^{2}x+\cos x)}{x\sin x\cos x}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^2\left(x/2\right)(1+{\cos }^{2}x+\cos x)}{2x\sin \left(x/2\right)\cos \left(x/2\right)\cos \left(x/2\right)\cos x}$

$=\underset{x\to 0}{lim}\frac{\sin \left(x/2\right)}{2.\left(x/2\right)}.\frac{1+{\cos }^{2}x+\cos x}{\cos \left(x/2\right)\cos x}$ $=\frac{1}{2}.1.\frac{3}{1}=\frac{3}{2}$