Question

Solve:

${lim}_{x\to 0}\frac{(1-\cos 3x)}{x\sin 2x}$

Answer 1

We have,

${lim}_{x\to 0}\frac{(1-\cos 3x)}{x\sin 2x}$

Apply L-hospital rule,

${lim}_{x\to 0}\frac{(0+3\sin 3x)}{\sin 2x+2x\cos 2x}$

${lim}_{x\to 0}\frac{\left(3\sin 3x\right)}{\sin 2x+2x\cos 2x}$

${lim}_{x\to 0}\frac{9x\left(\frac{\sin 3x}{3x}\right)}{2x\times \frac{\sin 2x}{2x}+2x\cos 2x}$

${lim}_{x\to 0}\frac{9\left(\frac{\sin 3x}{3x}\right)}{2\times \frac{\sin 2x}{2x}+2x\cos 2x}$

$=\frac{9}{2}$

Hence, this is the answer.