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. none of these

Answer 1

Answer: (B)

$\frac{2}{9}$

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

$={lim}_{x\to 0}f(2\frac{{\sin }^2\frac{3x}{2}}{{\left(\frac{3x}{2}\right)}^2}\times \frac{9}{4})$

$x\to 0$

$\frac{3x}{2}\to 0$

$\Rightarrow \frac{\sin \frac{3x}{2}}{\frac{3x}{2}}\to 1$

$={lim}_{x\to 0}f\left(\frac{2\times 1\times 9}{4}\right)={lim}_{x\to 0}f\left(\frac{9}{2}\right)$

$=\frac{2}{9}$