Chapter 12.2, Problem 12E

To determine

ToEvaluate:The limit $\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)$ , if it exists. Use a graphing utility to verify the result graphically.

Answer to Problem 12E

The $\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)=\frac{1}{4}$

Explanation of Solution

Given:

The $\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)$

Concept Used:

The limit of a function $f\left(x\right)$ at $x=c$ exists if the limit from right side of $c$ is equal to the limit from left side of $c$

  $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)$

Calculation:

For the $\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)$

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)=\underset{x\to 0}{\lim }\left(\left(\frac{\sqrt{x+4}-2}{x}\right)\left(\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\right)\right)\text{ Rationalising the denominator }\\ =\underset{x\to 0}{\lim }\left(\frac{{\left(\sqrt{x+4}\right)}^2-2^2}{x\left(\sqrt{x+4}+2\right)}\right)a^2-b^2=\left(a+b\right)\left(a-b\right)\\ =\underset{x\to 0}{\lim }\left(\frac{x+4-4}{x\left(\sqrt{x+4}+2\right)}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{x}{x\left(\sqrt{x+4}+2\right)}\right)\end{array}$

  $\begin{array}{c}=\underset{x\to 0}{\lim }\left(\frac{1}{1\left(\sqrt{x+4}+2\right)}\right)\\ =\frac{1}{\sqrt{0+4}+2}\text{ Substituting the limit}\\ =\frac{1}{2+2}\\ =\frac{1}{4}\end{array}$

Therefore, $\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)=\frac{1}{4}$

Using graphing utility, the graph of the function $f\left(x\right)=\frac{\sqrt{x+4}-2}{x}$ is

  

Since, as $x$ tends to $0$ , the graph of the function $f\left(x\right)=\frac{\sqrt{x+4}-2}{x}$ tends to $0.25$ .

Therefore, the graph of the function $f\left(x\right)=\frac{\sqrt{x+4}-2}{x}$ verifies the result.

Conclusion:

The $\underset{x\to 0}{\lim }\left(\frac{\sqrt{x+4}-2}{x}\right)=\frac{1}{4}$