Ulearngo

Types of Discontinuities

As we have seen in Example 2.26 and Example 2.27, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure 2.37 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.

Three graphs, each showing a different discontinuity. The first is removable discontinuity. Here, the given function is a line with positive slope. At a point x=a, where a>0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x<=a, and the second exists for x>a, where a>0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.
Figure 2.37 Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.

These three discontinuities are formally defined as follows:

Definition

If \(f \left(\right. x \left.\right)\) is discontinuous at a, then

  1. \(f\) has a removable discontinuity at a if \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right)\) exists. (Note: When we state that \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right)\) exists, we mean that \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = L ,\) where L is a real number.)
  2. \(f\) has a jump discontinuity at a if \(\underset{x \rightarrow a^{-}}{\text{lim}} f \left(\right. x \left.\right)\) and \(\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right)\) both exist, but \(\underset{x \rightarrow a^{-}}{\text{lim}} f \left(\right. x \left.\right) \neq \underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right) .\) (Note: When we state that \(\underset{x \rightarrow a^{-}}{\text{lim}} f \left(\right. x \left.\right)\) and \(\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right)\) both exist, we mean that both are real-valued and that neither take on the values ±∞.)
  3. \(f\) has an infinite discontinuity at a if \(\underset{x \rightarrow a^{-}}{\text{lim}} f \left(\right. x \left.\right) = \pm \infty\) and/or \(\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right) = \pm \infty .\)

Example 2.30

Classifying a Discontinuity

In Example 2.26, we showed that \(f \left(\right. x \left.\right) = \frac{x^{2} - 4}{x - 2}\) is discontinuous at \(x = 2 .\) Classify this discontinuity as removable, jump, or infinite.

Solution

To classify the discontinuity at 2 we must evaluate \(\underset{x \rightarrow 2}{\text{lim}} f \left(\right. x \left.\right) :\)

\[\begin{aligned} \underset{x \rightarrow 2}{\text{lim}} f \left(\right. x \left.\right) & = \underset{x \rightarrow 2}{\text{lim}} \frac{x^{2} - 4}{x - 2} \\ & = \underset{x \rightarrow 2}{\text{lim}} \frac{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)}{x - 2} \\ & = \underset{x \rightarrow 2}{\text{lim}} \left(\right. x + 2 \left.\right) \\ & = 4 . \end{aligned}\]

Since f is discontinuous at 2 and \(\underset{x \rightarrow 2}{\text{lim}} f \left(\right. x \left.\right)\) exists, f has a removable discontinuity at \(x = 2 .\)

Example 2.31

Classifying a Discontinuity

In Example 2.27, we showed that \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. - x^{2} + 4 & \text{if} x \leq 3 \\ 4 x - 8 & \text{if} x > 3 \end{aligned}\) is discontinuous at \(x = 3 .\) Classify this discontinuity as removable, jump, or infinite.

Solution

Earlier, we showed that f is discontinuous at 3 because \(\underset{x \rightarrow 3}{\text{lim}} f \left(\right. x \left.\right)\) does not exist. However, since \(\underset{x \rightarrow 3^{-}}{\text{lim}} f \left(\right. x \left.\right) = −5\) and \(\underset{x \rightarrow 3^{+}}{\text{lim}} f \left(\right. x \left.\right) = 4\) both exist, we conclude that the function has a jump discontinuity at 3.

Example 2.32

Classifying a Discontinuity

Determine whether \(f \left(\right. x \left.\right) = \frac{x + 2}{x + 1}\) is continuous at −1. If the function is discontinuous at −1, classify the discontinuity as removable, jump, or infinite.

Solution

The function value \(f \left(\right. −1 \left.\right)\) is undefined. Therefore, the function is not continuous at −1. To determine the type of discontinuity, we must determine the limit at −1. We see that \(\underset{x \rightarrow −1^{-}}{\text{lim}} \frac{x + 2}{x + 1} = − \infty\) and \(\underset{x \rightarrow −1^{+}}{\text{lim}} \frac{x + 2}{x + 1} = + \infty .\) Therefore, the function has an infinite discontinuity at −1.

Checkpoint 2.23

For \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. x^{2} & \text{if} x \neq 1 \\ 3 & \text{if} x = 1 , \end{aligned}\) decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite.

This lesson is part of:

Limits

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