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

$\frac{2}{9}$

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

$=f\left(\underset{x→0}{\lim}\frac{2\sin^2\frac{3x}{2}}{x^2}\right)=f\left(\underset{x→0}{\lim}2.\left(\frac{sin\frac{3x}{2}}{3x/2}\right).\frac{9}{4}\right)=f(\frac{9}{2})=\frac{2}{9}$

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