Question

${lim}_{x\to 0}\frac{1-\cos ax}{x\sin 2x}=$

• A. $\frac{a}{2}$
• B. $\frac{a^2}{2}$
• C. $\frac{a}{4}$
• D. $\frac{a^2}{4}$

Answer 1

The correct option is B $\frac{a^2}{2}$

Given,

$\begin{array}{c}{lim}_{x\to 0}\frac{1-\cos ax}{x\sin 2x}\\ ={lim}_{x\to 0}\frac{a\sin ax}{2x\cos 2x+\sin 2x}\\ ={lim}_{x\to 0}\frac{a^{2}x}{2x\cos 2x}\\ =\frac{a^2}{2}\end{array}$

Hence,

Option $B$ is correct.