Chapter 14.3, Problem 57AYU

$f\left(x\right)=\{\begin{array}{l}\frac{x^3-1}{x^2-1}ifx<1\\ 2ifx=1\\ \frac{3}{x+1}ifx>1\end{array}c=1$

To determine

To find: Whether $f$ is continuous at $c$.

Answer to Problem 57AYU

Not continuous.

Explanation of Solution

Given:

$f\left(x\right)=\frac{x^3-1}{x^2-1};x<1$

$f\left(x\right)=2;x=1$

$f\left(x\right)=\frac{3}{x+1};x>1$

$c=1$

Calculation:

We see that,

$\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(\frac{x^3-1}{x^2-1}\right)=\underset{x\to 1^{-}}{\lim }\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)}=\underset{x\to 1^{-}}{\lim }\frac{\left(x^2+x+1\right)}{\left(x+1\right)}=\frac{1^2+1+1}{1+1}=\frac{3}{2}$

Also,

$\underset{x\to c^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(\frac{3}{x+1}\right)=\underset{x\to 1^{+}}{\lim }\left(\frac{3}{1+1}\right)=\frac{3}{2}$

Again,

$f\left(c\right)=f\left(1\right)=2$; by definition.

Since $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)\ne f\left(c\right)$, therefore $f$ is not continuous at $c=1$.