Question

$\underset{x\to 0}{lim}\left(\frac{1-\cos 3x}{x^2}\right)$ is equal to

• A. $\frac{9}{2}$
• B. $0$
• C. $\frac{2}{9}$
• D. $\frac{8}{9}$

Answer 1

Answer: (A)

$\frac{9}{2}$

We have,

$\underset{x\to 0}{lim}\left(\frac{1-\cos 3x}{x^2}\right)$

This is $\frac{0}{0}$ form.

So, apply L-Hospital rule

$\underset{x\to 0}{lim}\left(\frac{0-3(-\sin 3x)}{2x}\right)$

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

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

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

$\frac{9}{2}\underset{x\to 0}{lim}\left(\frac{\sin 3x}{3x}\right)$

Since,

$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$

Therefore,

$=\frac{9}{2}\times 1$

$=\frac{9}{2}$

Hence, this is the answer.