Chapter 12, Problem 8RE

To determine

ToCalculate:The limit $\underset{x\to 2}{\lim }\left(\frac{1}{x-2}\right)$ using the given graph, if it exists. If the limit does not exist, explain why.

  

Answer to Problem 8RE

Since, $\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}\right)\ne \underset{x\to 2^{+}}{\lim }\left(\frac{1}{x-2}\right)$

Therefore, the given limit $\underset{x\to 2}{\lim }\left(\frac{1}{x-2}\right)$ does not exists.

Explanation of Solution

Given:

Thegraph of the function $f\left(x\right)=\frac{1}{x-2}$

  

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:

From the graph of the function $f\left(x\right)=\frac{1}{x-2}$

  

Left hand limit

From the graph of $f\left(x\right)=\frac{1}{x-2}$ , as $x$ approaches $2$ from left hand side, the graph of the function $f\left(x\right)=\frac{1}{x-2}$ approaches to $-\infty$ .

That is,

  $\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}\right)=-\infty$

Right hand limit

From the graph of $f\left(x\right)=\frac{1}{x-2}$ , as $x$ approaches $2$ from right hand side, the graph of the function $f\left(x\right)=\frac{1}{x-2}$ approaches to $\infty$ .

That is,

  $\underset{x\to 2^{+}}{\lim }\left(\frac{1}{x-2}\right)=\infty$

Since, $\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}\right)\ne \underset{x\to 2^{+}}{\lim }\left(\frac{1}{x-2}\right)$

Therefore, the given limit $\underset{x\to 2}{\lim }\left(\frac{1}{x-2}\right)$ does not exists.

Conclusion:

Since, $\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}\right)\ne \underset{x\to 2^{+}}{\lim }\left(\frac{1}{x-2}\right)$

Therefore, the given limit $\underset{x\to 2}{\lim }\left(\frac{1}{x-2}\right)$ does not exists.