Chapter 14.3, Problem 58AYU

To determine

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

Answer to Problem 58AYU

Not continuous.

Explanation of Solution

Given:

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

$f\left(x\right)=2;x=\text{ 2}$

$f\left(x\right)=\frac{x-4}{x-1};x>\text{ 2}$

$c=2$

Calculation:

We see that,

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

Also,

$\underset{x\to c^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\left(\frac{x-4}{x-1}\right)=\underset{x\to 2^{+}}{\lim }\left(\frac{2-4}{2-1}\right)=\frac{-2}{1}=-2$

Again,

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

Since $\underset{x\to c^{-}}{\lim }f\left(x\right)=f\left(c\right)\ne \underset{x\to c^{+}}{\lim }f\left(x\right)$, therefore $f$ is not continuous at $c=2$.