Chapter 12, Problem 32CLT

To determine

Toevaluate:Theevaluation of the limit $L=\underset{x\to 0}{\lim }\frac{\frac{1}{x-3}+\frac{1}{3}}{x}$ using graphing utility.

Answer to Problem 32CLT

The limit exists as $L=\underset{x\to 0}{\lim }\frac{\frac{1}{x-3}+\frac{1}{3}}{x}=-\frac{1}{9}$ .

Explanation of Solution

Given information:

The given limit is $L=\underset{x\to 0}{\lim }\frac{\frac{1}{x-3}+\frac{1}{3}}{x}$

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 $L=\underset{x\to 0}{\lim }\frac{\frac{1}{x-3}+\frac{1}{3}}{x}$

  $L=\underset{x\to 0}{\lim }\frac{\frac{1}{x-3}+\frac{1}{3}}{x}=\underset{x\to 0}{\lim }\frac{\frac{3+x-3}{3\left(x-3\right)}}{x}=$

  $L=\underset{x\to 0}{\lim }\frac{x}{3x\left(x-3\right)}$

  $L=\underset{x\to 0}{\lim }\frac{1}{3\left(x-3\right)}=\frac{1}{3\left(0-3\right)}=-\frac{1}{9}$

Thus, the limit $L=-\frac{1}{9}$

Verification of limit with graph:

The graph of function $f\left(x\right)=\frac{\frac{1}{x-3}+\frac{1}{3}}{x}$ and its solved function are shown in Figure-1 and Figure-2 respectively.

  

From the graph, we observe that when $x\to 0$ the graph becomes parallel to y -axis, so limit cannot be calculated. The limit of solved function is $L=-\frac{1}{9}$ =-0111.