Chapter 11, Problem 8RE

To determine

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

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

  

Answer to Problem 8RE

The given limit exists.

Explanation of Solution

Given:

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

  

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 $f\left(x\right)=\frac{x^2-1}{x-1}$

  

Left hand limit

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

That is,

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

Right hand limit

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

That is,

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

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

Therefore, the given limit $\underset{x\to 1}{\lim }\left(\frac{x^2-1}{x-1}\right)$ exists.

Conclusion:

The given limit $\underset{x\to 1}{\lim }\left(\frac{x^2-1}{x-1}\right)$ exists.