Chapter 14.3, Problem 73AYU

To determine

To find: The continuity of $f$ at $c=-1$ and $c=1$ and Graph $R$.

Answer to Problem 73AYU

Not continuous at $c=-1$ but continuous at $c=1$.

Explanation of Solution

Given:

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

Calculation:

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

$R\left(x\right)=\frac{\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}$

$R\left(x\right)=\frac{1}{x+1}$

At $c=-1$; we see that,

${\lim }_{x\to c}R\left(x\right)={\text{ lim}}_{x\to -1}\frac{1}{\left(x+1\right)}=\frac{1}{-1+1}=\frac{1}{0}$; does not exist.

Therefore, $R$ is not continuous at $c=-1$.

At $c=1$; we see that,

${\lim }_{x\to c}R\left(x\right)={\lim }_{x\to 1}\frac{1}{\left(x+1\right)}=\frac{1}{1+1}=\frac{1}{2}$

Also, $R\left(1\right)=\frac{1}{1+1}=\frac{1}{2}$

So $\lim {}_{x\to 1}R\left(x\right)=R\left(1\right)$

Therefore, $R$ is continuous at $c=1$.

Graph of $R$ is as follows: