Chapter 14.3, Problem 56AYU

To determine

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

Answer to Problem 56AYU

Continuous.

Explanation of Solution

Given:

$f\left(x\right)=\frac{x^2-6x}{x^2+6x};x\ne 0$

$f\left(x\right)=-1;x=0$

$c=0$

Calculation:

We see that,

$\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{x^2-6x}{x^2+6x}\right)=\underset{x\to 0}{\lim }\frac{x\left(x-6\right)}{x\left(x+6\right)}=\underset{x\to 0}{\lim }\frac{\left(x-6\right)}{\left(x+6\right)}=\frac{0-6}{0+6}=\frac{-6}{6}=-1$

Again,

$f\left(c\right)=f\left(0\right)=-1$; by definition.

Since $\underset{x\to c}{\lim }f\left(x\right)\text{ = }f\left(c\right)$, therefore $f$ is continuous at $c=0$.