Answer:
[tex]\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7} = \frac{1}{6}[/tex]
Step-by-step explanation:
[tex]\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7}[/tex]
[tex]\frac{\sqrt{x+2}-3 }{x-7}[/tex]
[tex]\frac{\frac{1}{2} (x+2)^{-\frac{1}{2} }}{1}[/tex] <-- Take derivative of numerator and denominator expressions
[tex]{\frac{1}{2} (x+2)^{-\frac{1}{2} }}[/tex] <-- Simplify
[tex]\frac{1}{2}(\frac{1}{\sqrt{x+2} })[/tex] <-- Rewrite
[tex]\frac{1}{2}(\frac{1}{\sqrt{7+2} })[/tex] <-- Use direct substitution and plug in the limit x=7
[tex]\frac{1}{2}(\frac{1}{\sqrt{9} })[/tex]
[tex]\frac{1}{2}(\frac{1}3 })[/tex]
[tex]\frac{1}{6}[/tex]
Therefore, [tex]\lim_{x \to 7} \frac{\sqrt{x+2}-3 }{x-7} = \frac{1}{6}[/tex]
