How to Remove 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

• How to Find Continuity at a Point

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.

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

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

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

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

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