Chapter 12, Problem 33CLT

To determine

Toevaluate:The evaluation of the givenlimit $L=\underset{x\to 0}{\lim }\frac{\sqrt{x+16}-4}{x}$ using graphing utility.

Answer to Problem 33CLT

The limit exists as $L=\underset{x\to 0}{\lim }\frac{\sqrt{x+16}-4}{x}=\frac{1}{4}$ .

Explanation of Solution

Given information:

The given limit is $L=\underset{x\to 0}{\lim }\frac{\sqrt{x+16}-4}{x}$ .

Formulaused:

Let f be a function $f:R\to R$  is defined on the real line and  a, L  ?  R . Then L is said the limit of function f , as  x  approaches  a ,and is written as $\underset{x\to a}{\lim }f\left(x\right)=L$ .The limit $\underset{x\to a}{\lim }f\left(x\right)=L$ is calculated by solving for finite value of $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)=L$ by manipulating to exist or by calculating $\underset{h\to 0}{\lim }f\left(a+h\right)=L=\underset{h\to 0}{\lim }f\left(a-h\right)$ .

Calculations:

The given limit is $L=\underset{x\to 0}{\lim }\frac{\sqrt{x+16}-4}{x}$ .

  $L=\underset{x\to 0}{\lim }\frac{\sqrt{x+16}-4}{x}\times \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}$

  $L=\underset{x\to 0}{\lim }\frac{{\left(\sqrt{x+16}\right)}^2-{\left(4\right)}^2}{x\left(\sqrt{x+16}+4\right)}$

  $L=\underset{x\to 0}{\lim }\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}$

  $L=\underset{x\to 0}{\lim }\frac{1}{\left(\sqrt{x+16}+4\right)}=\frac{1}{4}$

Thus, the limit $L=\frac{1}{4}$ .

Verification of limit with graph:

The graph of function $L=\underset{x\to 0}{\lim }\frac{\sqrt{x+16}-4}{x}$ and its solved function are shown in Figure(a) and Figure (b) respectively.

  

From the graph, we observe that when $x\to 0$ the limit is undefined. But the limit of solved function is $L=\frac{1}{4}=0.25$ .