Mathematics - Limits Question with Solution | TestHub

Given:$f\left(\frac{9}{2}\right)=\frac{2}{9}$and$f\left(x\right)$is continuous.

$\underset{x\to 0}{\lim }f\left(\frac{1-cos3x}{x^2}\right)=f\left(\underset{x\to 0}{\lim }\frac{1-cos3x}{x^2}\right)$

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


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

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

$=f\left(\frac{9}{2}\times 1^2\right)$$\left\{\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right\}$

$=f\left(\frac{9}{2}\right)$

$=\frac{2}{9}$