Discontinuities and Continuous Extensions, calculus lesson

A function can break in very different ways. A single point might be missing from an otherwise smooth curve, the graph might step up like a staircase, it might shoot off to infinity, or it might wiggle faster and faster without settling down. Telling these apart matters, because only the first kind can be repaired by changing or adding one value.

Key terms

Finite one-sided limits

Definition Let $f$ be defined on intervals $(c - r, c)$ and $(c, c + r)$ for some $r > 0$ . We say $f$ has finite one-sided limits at $c$ if both

L^- = \lim_{x \to c^-} f(x) \quad \text{and} \quad L^+ = \lim_{x \to c^+} f(x)

exist as real numbers. The two-sided limit $\lim_{x \to c} f(x)$ exists exactly when $L^-$ and $L^+$ both exist and are equal, and then it equals their common value.

We say $f$ has a discontinuity at $c$ when $f$ is defined on such intervals on both sides of $c$ and either $c$ is not in the domain of $f$ , or $c$ is in the domain and $\lim_{x \to c} f(x) = f(c)$ fails. This is wider than simply not being continuous at $c$ . Continuity, as defined in the previous lesson, is only asked about points of the domain, so $\frac{1}{x}$ is a continuous function. A discontinuity can also sit at a gap in the domain, so the same $\frac{1}{x}$ has a discontinuity at $0$ . The two statements do not clash, because one is about points of the domain and the other also covers the gaps between them. The types below sort discontinuities by how the one-sided limits behave.

In plain words Look at the left side and the right side separately and ask where each one is heading. Most of the sorting comes from those two numbers.

Example Consider

f(x) = \frac{\lvert x \rvert}{x}, \qquad x \neq 0.

Its values are $-1$ for $x < 0$ and $1$ for $x > 0$ . So $L^- = -1$ and $L^+ = 1$ . Both are finite, but they differ, so $\lim_{x \to 0} f(x)$ does not exist.

Removable discontinuity

Definition A discontinuity of $f$ at $c$ is removable if $\lim_{x \to c} f(x) = L$ exists as a finite number, and either $f(c)$ is undefined or $f(c) \neq L$ .

In plain words The graph has a hole or a single misplaced point, and everything around it points to one value. Redefining $f(c) = L$ fixes the problem.

Example The function

f(x) = \frac{x^2 - 1}{x - 1}

is undefined at $x = 1$ . For $x \neq 1$ it equals $x + 1$ , so $\lim_{x \to 1} f(x) = 2$ . The discontinuity at $1$ is removable.

Jump discontinuity

Definition A discontinuity of $f$ at $c$ is a jump discontinuity if both one-sided limits $L^-$ and $L^+$ exist as finite numbers and $L^- \neq L^+$ . The number $L^+ - L^-$ is the size of the jump.

In plain words The graph comes in at one height from the left and leaves at another height to the right. No single value at $c$ can join the two pieces.

Example The function $\frac{\lvert x \rvert}{x}$ from above has $L^- = -1$ and $L^+ = 1$ at $0$ , so it has a jump of size $2$ there. The floor function $\lfloor x \rfloor$ has a jump of size $1$ at every integer.

Infinite discontinuity

Definition A discontinuity of $f$ at $c$ is an infinite discontinuity if at least one of the one-sided limits $\lim_{x \to c^-} f(x)$ or $\lim_{x \to c^+} f(x)$ is $\infty$ or $-\infty$ . Then the line $x = c$ is a vertical asymptote of the graph.

In plain words The function blows up near $c$ . Its values get larger and larger in size, so no number can be assigned at $c$ to make it continuous.

Example For the function

f(x) = \frac{1}{(x - 2)^2}

both one-sided limits at $2$ are $\infty$ . At $x = 1.9$ the value is $100$ , at $x = 1.99$ it is $10{,}000$ . For $g(x) = \frac{1}{x}$ the left limit at $0$ is $-\infty$ and the right limit is $\infty$ , another infinite discontinuity.

Oscillating discontinuity

Definition A discontinuity of $f$ at $c$ is an oscillating discontinuity if at least one one-sided limit at $c$ does not exist as a real number and is not $\infty$ or $-\infty$ either. Then the values on that side never settle. For the right side, for example, there is a fixed $\varepsilon > 0$ such that every interval $(c, c + \delta)$ contains two points where the values of $f$ differ by at least $\varepsilon$ .

In plain words Close to $c$ the graph keeps going up and down without settling, no matter how far you zoom in.

Example Let $f(x) = \sin \frac{1}{x}$ for $x \neq 0$ . For every positive integer $k$ the points

a_k = \frac{2}{(4k + 1)\pi} \quad \text{and} \quad b_k = \frac{2}{(4k + 3)\pi}

give $f(a_k) = \sin\left(2k\pi + \frac{\pi}{2}\right) = 1$ and $f(b_k) = \sin\left(2k\pi + \frac{3\pi}{2}\right) = -1$ . Both lists of points approach $0$ from the right, so the values hit $1$ and $-1$ in every interval $(0, \delta)$ . The right limit does not exist, and the same happens from the left. The discontinuity at $0$ is oscillating. Every discontinuity is of at least one of the four types. A point can fit two of them when the function behaves one way on the left and another way on the right, for example blowing up on one side and oscillating on the other.

Piecewise functions

Definition A piecewise function is defined by different formulas on different parts of its domain. If each formula is continuous on the open piece where it is used, the only points that need checking are the break points. At a break point $c$ , compute the one-sided limits from the two neighboring formulas and compare them with $f(c)$ .

In plain words Inside each piece nothing new happens. All the action is at the seams.

Example Let

f(x) = \begin{cases} x^2, & x < 1, \\ 2x, & x \ge 1. \end{cases}

At $c = 1$ the left limit is $1^2 = 1$ and the right limit is $2 \cdot 1 = 2$ , with $f(1) = 2$ . The one-sided limits are finite and different, so $f$ has a jump of size $1$ at $x = 1$ . It is continuous from the right there but not from the left.

Continuous extension

Definition Let $f$ be continuous on its domain $D$ , let $c \notin D$ , and suppose $f$ is defined on an open interval around $c$ except at $c$ . If $\lim_{x \to c} f(x) = L$ exists as a finite number, the function

F(x) = \begin{cases} f(x), & x \in D, \\ L, & x = c \end{cases}

is continuous at every point of $D \cup \{c\}$ and agrees with $f$ on $D$ . It is called the continuous extension of $f$ to $c$ . If the limit does not exist as a finite number, no continuous extension to $c$ is possible.

In plain words When the only problem is a hole, fill the hole with the value the limit predicts.

Example Take

f(x) = \frac{x^2 + x - 6}{x^2 - 4},

which is undefined at $x = 2$ and $x = -2$ . For $x \neq \pm 2$ factoring gives

f(x) = \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)} = \frac{x + 3}{x + 2}.

At $2$ the limit is $\frac{5}{4}$ , so setting $F(2) = \frac{5}{4}$ extends $f$ continuously to $2$ . At $-2$ the numerator tends to $1$ while the denominator tends to $0$ , so the values blow up and there is no continuous extension to $-2$ .

Check yourself

Classify the discontinuity of $f(x) = \frac{x^2 - 9}{x + 3}$ at $x = -3$ and give the continuous extension if it exists.

Show the answer

For $x \neq -3$ the function equals $x - 3$ , so the limit at $-3$ is $-6$ . The discontinuity is removable, and $F(-3) = -6$ gives the continuous extension, which is simply $F(x) = x - 3$ .

Let $f(x) = x + 1$ for $x < 0$ and $f(x) = x^2 - 1$ for $x \ge 0$ . What kind of discontinuity is at $0$ , and how big is it?

Show the answer

The left limit is $1$ and the right limit is $-1$ . Both are finite and they differ, so it is a jump discontinuity of size $-1 - 1 = -2$ , a drop of $2$ .

Does $f(x) = \frac{x}{x^2 - x}$ have a continuous extension to $x = 0$ ? To $x = 1$ ?

Show the answer

For $x \neq 0, 1$ it equals $\frac{1}{x - 1}$ . At $0$ the limit is $-1$ , so $F(0) = -1$ works. At $1$ the values blow up, an infinite discontinuity, so no extension to $1$ exists.

Discontinuities and Continuous Extensions

Theory, part 29

Practice, part 30

Key terms

Finite one-sided limits

Removable discontinuity

Jump discontinuity

Infinite discontinuity

Oscillating discontinuity

Piecewise functions

Continuous extension

Check yourself

How this lesson connects

Builds on