# 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

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

## 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.

- \(\color{blue}{f(x)=\frac{x^3+64}{x+4}}\), \(\color{blue}{a=-4}\)
- \(\color{blue}{f(x)=\frac{2-\sqrt{x}}{4-x}}\), \(\color{blue}{a=4}\)
- \(\color{blue}{f(x)=\frac{5x^2+4x-1}{x+1}}\), \(\color{blue}{a=-1}\)
- \(\color{blue}{f(x)=\frac{9x^2-4}{3x+2}}\), \(\color{blue}{a=-\frac{2}{3}}\)

- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{x^3+64}{x+4} \quad if\:\ x ≠ -4 \\ 48 \quad if\:\ x=-4 \end{cases}}\)

- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{2-\sqrt{x}}{4-x} \quad if\:\ x ≠ 4 \\ \frac{1}{4} \quad if\:\ x=4 \end{cases}}\)

- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{5x^2+4x-1}{x+1} \quad if\:\ x ≠ -1 \\ -6 \quad if\:\ x=-1 \end{cases}}\)

- \(\color{blue}{g(x)=}\)\(\color{blue}{\begin{cases} \frac{9x^2-4}{3x+2} \quad if\:\ x ≠ -\frac{2}{3} \\ -4 \quad if\:\ x=-\frac{2}{3} \end{cases}}\)

### More math articles

- How Is the TSI Test Scored?
- How to Multiply Matrix? (+FREE Worksheet!)
- 6th Grade Common Core Math Practice Test Questions
- Overview of the HiSET Mathematics Test
- Top Calculator for the PSAT 10 Math Test: Quick Review
- How to Get a GED Certificate?
- Bеѕt Digital Pеn Tablet fоr Onlinе Mаth Teaching in 2022
- What Kind of Math Is on the ACT Test?
- ISEE Middle-Level Math Practice Test Questions
- Top 10 8th Grade NYSE Math Practice Questions

## What people say about "How to Remove Discontinuities?"?

No one replied yet.