Question

Pre-calc, image attached

Answer 1

Answer:

$\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7} = \frac{1}{6}$

Step-by-step explanation:

$\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7}$

$\frac{\sqrt{x+2}-3 }{x-7}$

$\frac{\frac{1}{2} (x+2)^{-\frac{1}{2} }}{1}$ <-- Take derivative of numerator and denominator expressions

${\frac{1}{2} (x+2)^{-\frac{1}{2} }}$ <-- Simplify

$\frac{1}{2}(\frac{1}{\sqrt{x+2} })$ <-- Rewrite

$\frac{1}{2}(\frac{1}{\sqrt{7+2} })$ <-- Use direct substitution and plug in the limit x=7

$\frac{1}{2}(\frac{1}{\sqrt{9} })$

$\frac{1}{2}(\frac{1}3 })$

$\frac{1}{6}$

Therefore, $\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7} = \frac{1}{6}$