Question

If $f\left(x\right)$ is continuous and $f\left(\frac{9}{2}\right)=\frac{2}{9}$, then $\underset{x\to 0}{lim}f\left(\frac{1-\cos 3x}{x^2}\right)$ is equal to

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

Answer 1

The correct option is B $\frac{2}{9}$

$L=\underset{x\to 0}{lim}\left(\frac{1-\cos 3x}{x^2}\right)$
$=\underset{x\to 0}{lim}\frac{2{\sin }^2\left(3x/2\right)}{x^2}$
$=\underset{x\to 0}{lim}2{\left(\frac{{\sin }^2\left(3x/2\right)}{3x/2}\right)}^2\times \frac{9}{4}$
$=2\times \frac{9}{4}$
$=\frac{9}{2}$

$\therefore \underset{x\to 0}{lim}f\left(\frac{1-\cos 3x}{x^2}\right)=f\left(\frac{9}{2}\right)=\frac{2}{9}$