How to Remove Discontinuities?
A removable discontinuity is a point on a graph that is not defined or does not match the rest of the graph. There is a gap in the chart at that location. Here you will learn more about removing discontinuities
![How to Remove Discontinuities?](https://www.effortlessmath.com/wp-content/uploads/2022/01/How-to-Remove-Discontinuities-512x240.jpg)
A discontinuity at \(c\) is called removable when the two-sided limit exists at \(c\) but isn’t equal to \(f(c)\). These types of discontinuities can be found by identifying values of the independent variable that are zeros for factors of the denominator that are also factors of the numerator of the given function. The function can be made continuous at these values by redefining \(f(c)\).
Related Topics
A step-by-step guide to removing discontinuities
If there is a discontinuity at \(x=c\) it is said to be removable if
\(lim_{x\to c}f(x)\) exists. Let’s call it \(L\).
However, \(L≠f (c)\) (Either because \(f(c)\) is a different number than \(L\) or because \(f(c)\) has not been defined.)
By introducing a new function, say \(g\), we can “delete” the discontinuity \((x)\):
\(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} f(x) \quad if\:\ x ≠ c \\ L \quad if\:\ x=c \end{cases}}\)
We now have \(g(x)=f(x)\) for all \(x≠c\) and \(g\) is continuous at \(c\).
Removing Discontinuities – Example 1:
The function \(f(x)=\frac{x^2-1}{x-1}\) is discontinuous at \(x=1\). Find what value should be assigned to \(f(1)\) to make\(f(x)\) continuous.
\(f(x)=\frac{x^2-1}{x-1}\) is discontinuous at \(x=1\) becuse \(f(1)\) does not exist.
But \(lim_{x\to 1}\frac{x^2-1}{x-1}= lim_{x\to 1}\frac{(x-1)(x+1)}{x-1}=x+1=1+1=2\)
So we remove the discontinuity by defining:
\(g(x)=\)\(\begin{cases} \frac{x^2-1}{x-1} \quad if\:\ x ≠ 1 \\ 2 \quad \ if\:\ x=1 \end{cases}\)
Removing Discontinuities – Example 2:
The function \(f(x)=\frac{sin x}{x}\) is discontinuous at \(x=0\). Find what value should be assigned to \(f(0)\) to make\(f(x)\) continuous.
The function is not defined in \(x=0\). As \(sin x\) is continuous at every \(x\), then the initial function \(f(x)=\frac{sin x}{x}\) is continuous for all \(x\) except the point \(x=0\).
\(g(x)=\)\(\begin{cases} \frac{sin x}{x} \quad if\:\ x ≠ 0 \\ 1 \quad \ if\:\ x=0 \end{cases}\)
Exercises for Removing Discontinuities
In each of the following functions, remove the discontinuity at the given point.
- \(\color{blue}{f(x)=\frac{x^3+64}{x+4}}\), \(\color{blue}{a=-4}\)
- \(\color{blue}{f(x)=\frac{2-\sqrt{x}}{4-x}}\), \(\color{blue}{a=4}\)
- \(\color{blue}{f(x)=\frac{5x^2+4x-1}{x+1}}\), \(\color{blue}{a=-1}\)
- \(\color{blue}{f(x)=\frac{9x^2-4}{3x+2}}\), \(\color{blue}{a=-\frac{2}{3}}\)
![This image has an empty alt attribute; its file name is answers.png](https://www.effortlessmath.com/wp-content/uploads/2019/12/answers.png)
- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{x^3+64}{x+4} \quad if\:\ x ≠ -4 \\ 48 \quad if\:\ x=-4 \end{cases}}\)
- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{2-\sqrt{x}}{4-x} \quad if\:\ x ≠ 4 \\ \frac{1}{4} \quad if\:\ x=4 \end{cases}}\)
- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{5x^2+4x-1}{x+1} \quad if\:\ x ≠ -1 \\ -6 \quad if\:\ x=-1 \end{cases}}\)
- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{9x^2-4}{3x+2} \quad if\:\ x ≠ -\frac{2}{3} \\ -4 \quad if\:\ x=-\frac{2}{3} \end{cases}}\)
Related to This Article
More math articles
- 4th Grade SBAC Math Worksheets: FREE & Printable
- 5th Grade MEA Math Worksheets: FREE & Printable
- What is the Side Splitter Theorem? A Complete Introduction and Exploration
- Ratio Tables
- 8th Grade MEAP Math Practice Test Questions
- Intelligent Math Puzzle – Challenge 87
- Getting a Math Degree: Hacks to Make Your Life Easier
- How to Order Fractions: Step-by-Step Guide
- Top 10 ACT Math Prep Books (Our 2023 Favorite Picks)
- How to Using Decimals, Grid Models, and Fractions to Represent Percent
What people say about "How to Remove Discontinuities? - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.