Let $f:\mathbf{R}\to [0,\mathrm{\infty }]$ be such that $\underset{x\to 3}{lim} f\left(x\right)$

exists and $\underset{x\to 3}{lim} \frac{\left(f\right(x){)}^2-4}{\sqrt{|x-3|}}=1$ Then $\underset{x\to 3}{lim} f\left(x\right)$ equal

Since, $\underset{x\to 3}{lim} \frac{\left(f\right(x){)}^2-4}{\sqrt{|x-3|}}=1$ So

for $ϵ=1$ there is $\delta >0$ such that

$0<\frac{\left(f(x{)}^2-9\right)}{\sqrt{|x-5|}}<2\text{ whenever }0<|x-3|<\delta$

$\Rightarrow 0<f\left(x\right)-2<2\frac{\sqrt{|x-3|}}{f\left(x\right)+2}$ whenever $0<|x-3|<\delta$

Since the range of $f$ is $[0,\mathrm{\infty }]$ so $\underset{x\to 3}{lim} \left(f\right(x)+2)\ne 0$ and exists. Thus 

$0\le \underset{x\to 3}{lim} \left(f\right(x)-2)\le 0\Rightarrow \underset{x\to 3}{lim} f\left(x\right)=2$