Question

${lim}_{x-0}\frac{1-{\cos }^{2}x}{\sin 3x\sin 5x}=$

• A. $\frac{1}{15}$
• B. $\frac{2}{15}$
• C. $\frac{1}{30}$
• D. $\frac{1}{10}$

Answer 1

The correct option is A $\frac{1}{15}$

We have,

$\begin{array}{c}{lim}_{x\to 0}\frac{1-{\cos }^{2}x}{\sin 3x\sin 5x}\\ ={lim}_{x\to 0}\frac{-2\cos x\left(-\sin x\right)}{3\cos 3x\sin 5x+5\cos 5x\sin 3x}\\ ={lim}_{x\to 0}\frac{2\cos 2x}{\left(3\cos 3x\right)\left(5\cos 5x\right)+\left(\sin 5x\right)\left(-9\sin x3x\right)+\left(5\cos 5x\right)\left(2\cos 3x\right)+\left(-25\sin 5x\right)\sin 3x}\\ =\frac{2}{15+15}=\frac{1}{15}\end{array}$

Hence, this is the answer.