Let $f : R\to [0, \infty )$be such that $\underset{x\to 5}{lim} f\left(x\right)$ exists and $\underset{x\to 5}{lim} \frac{\left(f\right(x){)}^2-9}{\sqrt{|x-5|}}=0.$ Then, $\underset{x\to 5}{lim} f\left(x\right)$ equals

Solution:

$\begin{array}{r}\underset{x\to 5}{lim} \frac{\left(f\right(x){)}^2-9}{\sqrt{|x-5|}}=0\\ \Rightarrow \underset{x\to 5}{lim} \left(f\right(x){)}^2-9=0\end{array}$

$\Rightarrow$ $l^2-9=0,$ where $\underset{x\to 5}{\lim } f\left(x\right)=l$

$\Rightarrow l=\pm 3$

$\Rightarrow l=3$                                          $[\therefore f(x)\ge 0$ for all $x \in R]$

$\Rightarrow \underset{x\to 5}{\lim } f\left(x\right)=3$