## Calculus: Early Transcendentals (2nd Edition)

$4$
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 --- $\displaystyle \lim_{x\rightarrow 5}\sqrt{x^{2}-9}=\lim_{x\rightarrow 5}(x^{2}-9)^{1/2}=$ ...apply the rule $=[\displaystyle \lim_{x\rightarrow 5}(x^{2}-9)]^{1/2}$=... ...since $x^{2}-9$ is a polynomial, by Th.2.4 $\displaystyle \lim_{x\rightarrow 5}(x^{2}-9)=5^{2}-9=16$ $...=[\displaystyle \lim_{x\rightarrow 5}(x^{2}-9)]^{1/2}=16^{1/2}=4$