Chapter 14.3, Problem 51AYU

$f\left(x\right)=\frac{x^3+3x}{x^2-3x}c=0$

To determine

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

Answer to Problem 51AYU

Not continuous.

Explanation of Solution

Given:

$f\left(x\right)=\frac{x^3+3x}{x^2-3x}$; $c=0$

Calculation:

We see that,

$\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{x^3+3x}{x^2-3x}\right)=\underset{x\to 0}{\lim }\frac{x\left(x^2+3\right)}{x\left(x-3\right)}=\underset{x\to 0}{\lim }\frac{\left(x^2+3\right)}{\left(x-3\right)}=\frac{0^2+3}{0-3}=\frac{3}{-3}=-1$

Again,

$f\left(c\right)=f\left(0\right)=\frac{0^3+3\left(0\right)}{0^2-3\left(0\right)}=\frac{0}{0}$; undefined.

Since $f\left(c\right)$ is undefined, therefore $f$ is not continuous at $c=0$.