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
Answer & explanation
Correct answer: option 2
$\underset{x→0}{\lim}f\left(\frac{1-\cos 3x}{x^2}\right)=f\left(\underset{x→0}{\lim}\frac{1-\cos 3x}{x^2}\right)$
$=f\left(\underset{x→0}{\lim}\frac{2\sin^2\frac{3x}{2}}{x^2}\right)=f\left(\underset{x→0}{\lim}2.\left(\frac{sin\frac{3x}{2}}{3x/2}\right).\frac{9}{4}\right)=f(\frac{9}{2})=\frac{2}{9}$
Hence $\underset{x→0}{\lim}\left(\frac{1-\cos 3x}{x^2}\right)=\frac{2}{9}$