Answer
$\lim_{x\to7}\dfrac{\sqrt{x+2}-3}{x-7}=\dfrac{1}{6}$
Work Step by Step
$\lim_{x\to7}\dfrac{\sqrt{x+2}-3}{x-7}$
Try to evaluate the limit using direct substitution:
$\lim_{x\to7}\dfrac{\sqrt{x+2}-3}{x-7}=\dfrac{\sqrt{7+2}-3}{7-7}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Rationalize the numerator and simplify:
$\lim_{x\to7}\dfrac{\sqrt{x+2}-3}{x-7}=\lim_{x\to7}\dfrac{\sqrt{x+2}-3}{x-7}\cdot\dfrac{\sqrt{x+2}+3}{\sqrt{x+2}+3}=...$
$...=\lim_{x\to7}\dfrac{(\sqrt{x+2})^{2}-3^{2}}{(x-7)(\sqrt{x+2}+3)}=...$
$...=\lim_{x\to7}\dfrac{x+2-9}{(x-7)(\sqrt{x+2}+3)}=...$
$...=\lim_{x\to7}\dfrac{x-7}{(x-7)(\sqrt{x+2}+3)}=\lim_{x\to7}\dfrac{1}{\sqrt{x+2}+3}=...$
Try to evaluate the limit using direct substitution again:
$\lim_{x\to7}\dfrac{1}{\sqrt{x+2}+3}=\dfrac{1}{\sqrt{7+2}+3}=\dfrac{1}{3+3}=\dfrac{1}{6}$
