Answer
$4$
Work Step by Step
Theorem 2.3.7. Fractional power
$\displaystyle \lim_{x\rightarrow a}(f(x))^{n/m}=\left(\lim_{x\rightarrow a}f(x)\right)^{n/m},$
provided $f(x)\geq 0,$ for $x$ near $a,$ if $m$ is even and $n/m$ is reduced to lowest terms
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$\displaystyle \lim_{x\rightarrow 5}\sqrt{x^{2}-9}=\lim_{x\rightarrow 5}(x^{2}-9)^{1/2}=$
...apply the rule
$=[\displaystyle \lim_{x\rightarrow 5}(x^{2}-9)]^{1/2}$=...
...since $x^{2}-9$ is a polynomial, by Th.2.4
$\displaystyle \lim_{x\rightarrow 5}(x^{2}-9)=5^{2}-9=16$
$...=[\displaystyle \lim_{x\rightarrow 5}(x^{2}-9)]^{1/2}=16^{1/2}=4$
