12.3 Continuity  (Page 3/10)

1. Notice that the function is defined everywhere except at $x=5.$

   Thus, $f\left(5\right)$ does not exist, Condition 2 is not satisfied. Since Condition 1 is satisfied, the limit as $x$ approaches 5 is 8, and Condition 2 is not satisfied.This means there is a removable discontinuity at $x=5.$

2. Condition 2 is satisfied because $g(2)=-2.$

   Notice that the function is a piecewise function    , and for each piece, the function is defined everywhere on its domain. Let’s examine Condition 1 by determining the left- and right-hand limits as $x$ approaches 2.

   Left-hand limit: $\underset{x\to 2^{-}}{\lim }\left(x+1\right)=2+1=3.$ The left-hand limit exists.

   Right-hand limit: $\underset{x\to 2^{+}}{\lim }\left(-x\right)=-2.$ The right-hand limit exists. But

   $\underset{x\to 2^{-}}{\lim }f(x)\ne \underset{x\to 2^{+}}{\lim }f(x).$

   So, $\underset{x\to 2}{\lim }f(x)$ does not exist, and Condition 2 fails: There is no removable discontinuity. However, since both left- and right-hand limits exist but are not equal, the conditions are satisfied for a jump discontinuity at $x=2.$

To determine if the function $f$ is continuous at $x=a,$ we will determine if the three conditions of continuity are satisfied at $x=a$ .

1. Condition 1: Does $f(a)$ exist?

   $$\begin{array}{l}f(3)=4(3)=12\hfill \\ \Rightarrow \text{Condition 1 is satisfied}.\hfill \end{array}$$

   Condition 2: Does $\underset{x\to 3}{\lim }f(x)$ exist?

   To the left of $x=3,$ $f(x)=4x;$ to the right of $x=3,$ $f(x)=8+x.$ We need to evaluate the left- and right-hand limits as $x$ approaches 1.

   • Left-hand limit: $\underset{x\to 3^{-}}{\lim }f(x)=\underset{x\to 3^{-}}{\lim }4(3)=12$
   • Right-hand limit: $\underset{x\to 3^{+}}{\lim }f(x)=\underset{x\to 3^{+}}{\lim }\left(8+x\right)=8+3=11$

   Because $\underset{x\to 1^{-}}{\lim }f(x)\ne \underset{x\to 1^{+}}{\lim }f(x),$ $\underset{x\to 1}{\lim }f(x)$ does not exist.

   $\Rightarrow \text{ Condition 2 fails}\text{.}$

   There is no need to proceed further. Condition 2 fails at $x=3.$ If any of the conditions of continuity are not satisfied at $x=3,$ the function $f\left(x\right)$ is not continuous at $x=3.$

2. $x=\frac{8}{3}$

   Condition 1: Does $f\left(\frac{8}{3}\right)$ exist?

   $\begin{array}{l}f\left(\frac{8}{3}\right)=4\left(\frac{8}{3}\right)=\frac{32}{3}\hfill \\ \Rightarrow \text{Condition 1 is satisfied}.\hfill \end{array}$

   Condition 2: Does $\underset{x\to \frac{8}{3}}{\lim }f(x)$ exist?

   To the left of $x=\frac{8}{3},$ $f(x)=4x;$ to the right of $x=\frac{8}{3},$ $f(x)=8+x.$ We need to evaluate the left- and right-hand limits as $x$ approaches $\frac{8}{3}.$

   • Left-hand limit: $\underset{x\to {\frac{8}{3}}^{-}}{\lim }f(x)=\underset{x\to {\frac{8}{3}}^{-}}{\lim }4\left(\frac{8}{3}\right)=\frac{32}{3}$
   • Right-hand limit: $\underset{x\to {\frac{8}{3}}^{+}}{\lim }f(x)=\underset{x\to {\frac{8}{3}}^{+}}{\lim }\left(8+x\right)=8+\frac{8}{3}=\frac{32}{3}$

   Because $\underset{x\to \frac{8}{3}}{\lim }f(x)$ exists,

   $\Rightarrow \text{Condition 2 is satisfied}.$

   Condition 3: Is $f\left(\frac{8}{3}\right)=\underset{x\to \frac{8}{3}}{\lim }f(x)?$

   $\begin{array}{l}f\left(\frac{32}{3}\right)=\frac{32}{3}=\underset{x\to \frac{8}{3}}{\lim }f(x)\hfill \\ \Rightarrow \text{Condition 3 is satisfied}.\hfill \end{array}$

   Because all three conditions of continuity are satisfied at $x=\frac{8}{3},$ the function $f\left(x\right)$ is continuous at $x=\frac{8}{3}.$