Question

$\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin 2x}$ equals :-

• A. $1/4$
• B. $1$
• C. $3/4$
• D. Does not exist

Answer 1

The correct option is C $3/4$

$\begin{array}{c}{lim}_{x\to 0}\frac{1-{\cos }^{3}x}{x\sin 2x}\\ Here,\\ \left(1-{\cos }^{3}x\right)=\left(1-\cos x\right)\left(1+{\cos }^{2}x+\cos x\right)\\ =2{\sin }^2\frac{x}{2}\left(1+{\cos }^{2}x+\cos x\right)\\ andusetheformula\\ {lim}_{x\to 0}\frac{\sin x}{x}=1\\ Now,\\ {lim}_{x\to 0}\frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}\times \frac{x^2}{4}\frac{\left(1+{\cos }^{2}x+\cos x\right)}{x\left(\sin \frac{2x}{2x}\right)}\times 2x\\ =\left(2\times \frac{1}{4}\right)\frac{\left(1+1+1\right)}{2}\\ =\frac{3}{4}\end{array}$