do direct substitution first.
that gets you 0/0, which means that further work can be done
since theres a radical there, we can multiply the numerator and denominator by the conjugate of the part containing the square root, in this case the numerator
\(lim_{x\rightarrow 5}\frac{2-\sqrt{9-x}}{x-5}*\frac{2+\sqrt{9-x}}{2+\sqrt{9-x}}\\ =lim_{x\rightarrow 5}\frac{9-(9-x)}{(x-5)(2+\sqrt{9-x})}\\ =lim_{x\rightarrow 5}\frac{x-5}{(x-5)(2+\sqrt{9-x})}\\ =lim_{x\rightarrow 5}\frac{1}{2+\sqrt{9-x}}\\ =\frac{1}{2+\sqrt{9-5}}=\frac{1}{4}\)
