Recognizing Continuous and Discontinuous Real-Number Functions
Many of the functions we have encountered in earlier chapters are continuous everywhere. They never have a hole in them, and they never jump from one value to the next. For all of these functions, the limit of [latex]f\left(x\right)[/latex] as [latex]x[/latex] approaches [latex]a[/latex] is the same as the value of [latex]f\left(x\right)[/latex] when [latex]x=a[/latex]. So [latex]\underset{x\to a}{\mathrm{lim}}f\left(x\right)=f\left(a\right)[/latex]. There are some functions that are continuous everywhere and some that are only continuous where they are defined on their domain because they are not defined for all real numbers.
continuous functions
The following functions are continuous everywhere:
How To: Given a function [latex]\begin{align}f\left(x\right)\end{align}[/latex], determine if the function is continuous at [latex]\begin{align}x=a\end{align}[/latex].
If all three conditions are satisfied, the function is continuous at [latex]x=a[/latex]. If any one of the conditions is not satisfied, the function is not continuous at [latex]x=a[/latex].
Determine whether the function [latex]f\left(x\right)=\begin{cases}4x, \hfill& x\leq 3 \\ 8+x, \hfill& x>3\end{cases}[/latex] is continuous at
[latex]x=3[/latex]
[latex]x=\frac{8}{3}[/latex]
To determine if the function [latex]f[/latex] is continuous at [latex]x=a[/latex], we will determine if the three conditions of continuity are satisfied at [latex]x=a[/latex] .
Condition 1: Does [latex]f\left(a\right)[/latex] exist?
[latex]\begin{align}&f\left(3\right)=4\left(3\right)=12 \\ &\Rightarrow \text{Condition 1 is satisfied}. \end{align}[/latex]
Condition 2: Does [latex]\underset{x\to 3}{\mathrm{lim}}f\left(x\right)[/latex] exist?
To the left of [latex]x=3[/latex], [latex]f\left(x\right)=4x[/latex]; to the right of [latex]x=3[/latex], [latex]f\left(x\right)=8+x[/latex]. We need to evaluate the left- and right-hand limits as [latex]x[/latex] approaches 1.
Because [latex]\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)[/latex], [latex]\underset{x\to 1}{\mathrm{lim}}f\left(x\right)[/latex] does not exist.
There is no need to proceed further. Condition 2 fails at [latex]x=3[/latex]. If any of the conditions of continuity are not satisfied at [latex]x=3[/latex], the function [latex]f\left(x\right)[/latex] is not continuous at [latex]x=3[/latex].
[latex]x=\frac{8}{3}[/latex]Condition 1: Does [latex]f\left(\frac{8}{3}\right)[/latex] exist?
[latex]\begin{align}&f\left(\frac{8}{3}\right)=4\left(\frac{8}{3}\right)=\frac{32}{3} \\ &\Rightarrow \text{Condition 1 is satisfied}. \end{align}[/latex]
Condition 2: Does [latex]\underset{x\to \frac{8}{3}}{\mathrm{lim}}f\left(x\right)[/latex] exist?
To the left of [latex]x=\frac{8}{3}[/latex], [latex]f\left(x\right)=4x[/latex]; to the right of [latex]x=\frac{8}{3}[/latex], [latex]f\left(x\right)=8+x[/latex]. We need to evaluate the left- and right-hand limits as [latex]x[/latex] approaches [latex]\frac{8}{3}[/latex].
Because [latex]\underset{x\to \frac{8}{3}}{\mathrm{lim}}f\left(x\right)[/latex] exists,
[latex]\Rightarrow \text{Condition 2 is satisfied}[/latex].
Condition 3: Is [latex]f\left(\frac{8}{3}\right)=\underset{x\to \frac{8}{3}}{\mathrm{lim}}f\left(x\right)?[/latex]
[latex]\begin{align}f&\left(\frac{32}{3}\right)=\frac{32}{3}=\underset{x\to \frac{8}{3}}{\mathrm{lim}}f\left(x\right) \\ &\Rightarrow \text{Condition 3 is satisfied}. \end{align}[/latex]
Because all three conditions of continuity are satisfied at [latex]x=\frac{8}{3}[/latex], the function [latex]f\left(x\right)[/latex] is continuous at [latex]x=\frac{8}{3}[/latex].
Determine whether the function [latex]f\left(x\right)=\begin{cases}\frac{1}{x}, \hfill& x\leq 2 \\ 9x-17.5, \hfill& x>2\end{cases}[/latex] is continuous at [latex]x=2[/latex].
yes
Determine whether the function [latex]f\left(x\right)=\frac{{x}^{2}-25}{x - 5}[/latex] is continuous at [latex]x=5[/latex].
To determine if the function [latex]f[/latex] is continuous at [latex]x=5[/latex], we will determine if the three conditions of continuity are satisfied at [latex]x=5[/latex].
Condition 1:
[latex]\begin{gathered}f\left(5\right)\text{ does not exist.} \\ \Rightarrow \text{Condition 1 fails}.\end{gathered}[/latex]
There is no need to proceed further. Condition 2 fails at [latex]x=5[/latex]. If any of the conditions of continuity are not satisfied at [latex]x=5[/latex], the function [latex]f[/latex] is not continuous at [latex]x=5[/latex].
At [latex]x=5[/latex], there exists a removable discontinuity.
Determine whether the function [latex]f\left(x\right)=\frac{9-{x}^{2}}{{x}^{2}-3x}[/latex] is continuous at [latex]x=3[/latex]. If not, state the type of discontinuity.
No, the function is not continuous at [latex]x=3[/latex]. There exists a removable discontinuity at [latex]x=3[/latex].
