Chapter 13.2, Problem 19E

To determine

To evaluate the limit it exist

Answer to Problem 19E

The solution is 6

Explanation of Solution

Given information:

Given equation is,

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

CALCULATION:

To find the limit,

  $\underset{x\to a}{\lim f\left(x\right)}=L$

“The limit of $f\left(x\right)$ as $x$ approaches a, equals $L\text{'}$ .

The limit is obtained as:

  $\begin{array}{l}\underset{x\to 7}{\lim }\frac{\sqrt{x+2}-3}{x-7}\\ =\underset{x\to 7}{\lim }\frac{\sqrt{x+2}-3}{x-7}\times \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}\left[\text{rationalizing numerator}\right]\\ =\underset{x\to 7}{\lim }\frac{\left(x+2\right)-9}{\left(x-7\right)\left(\sqrt{x+2}+3\right)}\left[{\text{apply, a}}^2-b^2=\left(a-b\right)\left(a+b\right)\right]\\ \underset{x\to 7}{\lim }\frac{x-7}{\left(x-7\right)\left(\sqrt{x+2}+3\right)}\\ =\underset{x\to 7}{\lim }\frac{1}{\left(\sqrt{x+2}+3\right)}\text{}\left[\text{cancel,}\left(x-7\right)\right]\\ =\frac{1}{\sqrt{7+2}+3}\left[\text{substitute,}x=7\right]\\ =\frac{1}{\sqrt{9+3}}\\ =\frac{1}{3+3}=\frac{1}{6}\end{array}$

Hence,

The solution is 6.