Question

Evaluate : $\underset{x\to 0}{lim}\frac{{\sin }^{2}5x}{1-\cos x}$

• A. 5
• B. 50
• C. 1
• D. None of these

Answer 1

The correct option is B 50

We have, $\underset{x\to 0}{lim}\frac{{\sin }^{2}5x}{1-\cos x}$

Let $L=\underset{x\to 0}{lim}\frac{{\sin }^{2}5x}{1-\cos x}$

This is $\frac{0}{0}$ form, so apply L-hospital rule

$\underset{x\to 0}{lim}\frac{2\sin 5x\cos 5x\times 5}{0+\sin x}$

$L=\underset{x\to 0}{lim}\frac{5\left(2\sin 5x\cos 5x\right)}{\sin x}$

$=\underset{x\to 0}{lim}\frac{5\left(\sin 10x\right)}{\sin x}$

$=\underset{x\to 0}{lim}\frac{5\times \frac{\sin 10x}{10x}\times 10x}{\frac{\sin x}{x}\times x}$

$=\underset{x\to 0}{lim}\frac{50\times \frac{\sin 10x}{10x}}{\frac{\sin x}{x}}$

Since, ${lim}_{x\to 0}\frac{\sin x}{x}=1$

Therefore,

$L=\frac{50\times 1}{1}$

$L=50$

Hence, this is the answer.