Answer
$\lim\limits_{x \to 4}\dfrac{\sqrt{x+5}-3}{x-4}=\dfrac{1}{6}$
Work Step by Step
$\lim\limits_{x \to 4}\dfrac{\sqrt{x+5}-3}{x-4}$
Applying direct substitution immediately will result in an indeterminate form.
Multiply the fraction by $\dfrac{\sqrt{x+5}+3}{\sqrt{x+5}+3}$ and simplify:
$\lim\limits_{x \to 4}\Big(\dfrac{\sqrt{x+5}-3}{x-4}\Big)\Big(\dfrac{\sqrt{x+5}+3}{\sqrt{x+5}+3}\Big)=\lim\limits_{x \to 4}\dfrac{(\sqrt{x+5})^{2}-3^{2}}{(x-4)(\sqrt{x+5}+3)}$
$...=\lim\limits_{x \to 4}\dfrac{x+5-9}{(x-4)(\sqrt{x+5}+3)}=\lim\limits_{x \to 4}\dfrac{x-4}{(x-4)(\sqrt{x+5}+3)}=...$
$...=\lim\limits_{x \to 4}\dfrac{1}{\sqrt{x+5}+3}$
Now, apply direct substitution to evaluate the limit:
$\lim\limits_{x \to 4}\dfrac{1}{\sqrt{x+5}+3}=\dfrac{1}{\sqrt{4+5}+3}=\dfrac{1}{3+3}=\dfrac{1}{6}$
