As $\underset{x\to 5}{\lim }\frac{\left(f\right(x){)}^2-9}{\sqrt{|x-5|}}=0,\text{ so }$

$\left(f\right(x){)}^2-9=K|x-5{|}^{\alpha }P\left(x\right)\text{, where }\alpha >1/2\text{ and }P\left(x\right)$ is a

function of x such that $\underset{x\to 5}{\lim }P\left(x\right)$ exist

$\varphi \mathrm{l}\underset{x}{\lim }\left(f\right(x){)}^2=\underset{\{}{\lim }\left[K|x-5{|}^{\alpha }P(x)+9\right]=9$

The only possible answer is $\underset{x\to 5}{\lim }f\left(x\right)=3$