Continuity: 3 Conditions & Discontinuities

A function is continuous at x = a when its graph has no break there. In calculus, continuity at a requires three conditions: f(a) is defined, $\lim_{x \to a} f(x)$ exists, and the limit equals f(a). If any condition fails, f has a discontinuity at a: removable, jump or infinite.

Before this lesson: One-Sided Limits, Evaluating Limits Algebraically, Infinite Limits and Piecewise Functions

What continuity means

Informally, a continuous function is one you can draw without lifting your pencil. The formal definition turns that picture into something you can check with limits: the value the graph is heading toward must be the value it actually takes.

Most familiar functions are continuous everywhere they are defined: polynomials, rational functions (except where the denominator is 0), roots, exponentials, logarithms, sine and cosine. Breaks usually happen where a denominator is 0 or where a piecewise function changes rules.

Continuity matters because many calculus theorems, such as the intermediate value theorem in the next lesson, only work for continuous functions.

Three small graphs. Removable: a line with an open circle at one point, where the limit exists but the function is undefined or has a different value. Jump: the graph breaks and restarts at a different height, so the left and right limits differ. Infinite: the graph shoots up and down beside a dashed vertical asymptote. — The three kinds of discontinuity. Only the removable kind can be fixed by changing or defining one value.

The definition of continuity

A function f is continuous at x = a when all three conditions hold:

f(a) is defined.
$\lim_{x \to a} f(x)$ exists (the left- and right-hand limits are equal).
$\lim_{x \to a} f(x) = f(a)$ .

Type of discontinuity	What fails	How it looks
Removable	The limit exists, but f(a) is undefined or not equal to it	A hole, possibly with a dot elsewhere
Jump	The one-sided limits exist but are different	The graph breaks and restarts at a new height
Infinite	A one-sided limit is $\infty$ or $-\infty$	A vertical asymptote

A function is continuous on an interval if it is continuous at every point in it. At a closed endpoint, only the one-sided limit from inside the interval is needed.

Continuity examples

Example 1: A removable discontinuity

Is $f(x) = \frac{x^2 - 1}{x - 1}$ continuous at x = 1? If not, classify the discontinuity.

f(1) = \frac{0}{0}

Condition 1: substituting gives 0/0, so f(1) is undefined. f is not continuous at 1.

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x + 1)(x - 1)}{x - 1}

Check whether the limit exists: factor the top.

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2

Cancel and substitute (see evaluating limits algebraically).

Answer: not continuous at x = 1, but the limit exists, so the discontinuity is removable. Defining f(1) = 2 would make f continuous there.

Example 2: A piecewise function that is continuous

Let $f(x) = x^2 + 1$ for $x \lt 2$ and $f(x) = 3x - 1$ for $x \ge 2$ . Is f continuous at x = 2?

f(2) = 3(2) - 1 = 5

Condition 1: x = 2 uses the second piece. f(2) is defined.

\lim_{x \to 2^-} f(x) = 2^2 + 1 = 5

Left-hand limit: use the first piece.

\lim_{x \to 2^+} f(x) = 3(2) - 1 = 5

Right-hand limit: use the second piece.

\lim_{x \to 2} f(x) = 5 = f(2)

Conditions 2 and 3: the limit exists and equals f(2).

Answer: yes, f is continuous at x = 2. The two pieces meet at (2, 5).

Example 3: Classify every discontinuity

Find and classify the discontinuities of $g(x) = \frac{x + 3}{x^2 - 9}$ .

x^2 - 9 = (x - 3)(x + 3)

Factor the denominator. g is undefined at x = 3 and x = −3.

g(x) = \frac{1}{x - 3}

Cancel x + 3, for $x \ne -3$ .

\lim_{x \to -3} g(x) = \frac{1}{-3 - 3} = -\frac{1}{6}

At x = −3 the limit exists, so this discontinuity is removable.

\lim_{x \to 3^+} g(x) = \infty

At x = 3, 1 over a tiny positive number blows up (see infinite limits).

Answer: a removable discontinuity at x = −3 (a hole at $\left(-3, -\frac{1}{6}\right)$ ) and an infinite discontinuity at x = 3 (a vertical asymptote). g is continuous everywhere else.

Example 4: Find a constant that makes a function continuous

Find k so that $f(x) = kx^2$ for $x \le 3$ and $f(x) = 2x + k$ for $x \gt 3$ is continuous everywhere.

\lim_{x \to 3^-} f(x) = k(3)^2 = 9k

The left-hand limit (and f(3)) comes from the first piece.

\lim_{x \to 3^+} f(x) = 2(3) + k = 6 + k

The right-hand limit comes from the second piece.

9k = 6 + k

For continuity at 3, the two must be equal.

8k = 6

Subtract k from both sides (see equations with variables on both sides).

k = \frac{3}{4}

Divide both sides by 8 and simplify.

Answer: $k = \frac{3}{4}$ . Check: $\frac{3}{4}(9) = 6.75$ and $6 + 0.75 = 6.75$ . Both pieces are polynomials, so f is continuous at every other x as well.

Example 5: Word problem with a phone plan

A phone plan costs $40 per month for up to 10 GB of data. Above 10 GB it costs $55 + 5(g - 10)$ dollars for g GB. Let C(g) be the monthly cost. Is C continuous at g = 10?

C(10) = 40

Exactly 10 GB is in the first tier.

\lim_{g \to 10^-} C(g) = 40

Just under 10 GB, the cost is $40.

\lim_{g \to 10^+} C(g) = 55 + 5(10 - 10) = 55

Just over 10 GB, use the second rule.

40 \ne 55

The one-sided limits are different.

Answer: no. C has a jump discontinuity at 10 GB: using a tiny bit more data than 10 GB raises the bill by $15 all at once.

Common mistakes

Mistake	Example of the error	Fix
Checking only that f(a) exists	Calling the phone plan continuous because C(10) = 40	All three conditions are needed; here the one-sided limits differ.
Calling every hole an asymptote	Saying g has an asymptote at x = −3 in Example 3	If the factor cancels and the limit exists, it is removable.
Using the wrong piece at the boundary	Using $x^2 + 1$ for f(2) in Example 2	Check the inequality: $x \ge 2$ means x = 2 uses the second piece.
Thinking a limit that exists means continuous	$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$ , so “continuous”	f(1) must also be defined and equal 2.
Setting the pieces equal to each other instead of their limits	Solving $kx^2 = 2x + k$ for x	Substitute the boundary value x = 3 into both pieces first.
Forgetting domain limits	Calling $\sqrt{x - 3}$ discontinuous at 3	At an endpoint, only the one-sided limit from inside the domain matters; it is continuous on $[3, \infty)$ .

Related methods

Continuity of combinations

Sums, differences, products and compositions of continuous functions are continuous. A quotient is continuous wherever its denominator is not 0. So to find where a formula is continuous, find where it is undefined.

Limits of continuous functions

For a continuous function, direct substitution always gives the limit. That is why the limit laws let you substitute into polynomials, and why $\lim_{x \to 0} \cos x = \cos 0 = 1$ .

Practice problems

Is $f(x) = x^3 - 4x + 1$ continuous at x = 2? Explain.
Where is $f(x) = \frac{1}{x - 5}$ discontinuous, and what type is it?
Classify the discontinuity of $h(x) = \frac{x^2 - 4}{x + 2}$ at x = −2.
Is f(x) = 2x + 1 for $x \lt 1$ and f(x) = 4 − x for $x \ge 1$ continuous at x = 1?
Classify the discontinuity of f(x) = $x^2$ for $x \lt 0$ and f(x) = x + 1 for $x \ge 0$ at x = 0.
Find a so that f(x) = ax + 3 for $x \lt 2$ and f(x) = $x^2 + a$ for $x \ge 2$ is continuous.
On what interval is $f(x) = \sqrt{x - 3}$ continuous?
Where is $g(x) = \frac{x + 1}{x^2 - 5x + 6}$ continuous?
What value should $f(3)$ have so that $f(x) = \frac{x^2 - 7x + 12}{x - 3}$ is continuous at 3?
A ride costs C(m) = 4 + 1.5m dollars for m ≤ 20 miles and C(m) = 34 + 1.25(m − 20) for m > 20. Is C continuous at m = 20?

Show answers

f(2) = 8 - 8 + 1 = 1

Find f(2).

\lim_{x \to 2} f(x) = 1

A polynomial’s limit is found by substitution.

Answer: yes; f(2) = 1 equals the limit. Polynomials are continuous everywhere.

x - 5 = 0

The denominator is 0 at x = 5, and the numerator is 1, not 0.

\lim_{x \to 5^+} \frac{1}{x - 5} = \infty

1 over a tiny positive number.

Answer: discontinuous at x = 5; an infinite discontinuity (vertical asymptote).

\lim_{x \to -2} \frac{(x + 2)(x - 2)}{x + 2} = \lim_{x \to -2} (x - 2) = -4

Factor, cancel and substitute.

Answer: h(−2) is undefined but the limit is −4, so it is a removable discontinuity (a hole at (−2, −4)).

f(1) = 4 - 1 = 3

x = 1 uses the second piece.

\lim_{x \to 1^-} f(x) = 2(1) + 1 = 3

Left-hand limit.

\lim_{x \to 1^+} f(x) = 4 - 1 = 3

Right-hand limit.

Answer: yes; all three equal 3.

\lim_{x \to 0^-} f(x) = 0^2 = 0

Left-hand limit.

\lim_{x \to 0^+} f(x) = 0 + 1 = 1

Right-hand limit.

Answer: a jump discontinuity, since the one-sided limits (0 and 1) differ.

2a + 3 = 2^2 + a

Set the left-hand limit equal to the value from the second piece at x = 2.

2a + 3 = 4 + a

Simplify.

a = 1

Subtract a and 3 from both sides.

Answer: a = 1

x - 3 \ge 0

The root is defined when the inside is not negative.

x \ge 3

Add 3 to both sides.

Answer: continuous on $[3, \infty)$ .

x^2 - 5x + 6 = (x - 2)(x - 3)

Factor the denominator to find where g is undefined.

Answer: continuous for all real numbers except x = 2 and x = 3.

\frac{x^2 - 7x + 12}{x - 3} = \frac{(x - 3)(x - 4)}{x - 3} = x - 4

Factor and cancel, for $x \ne 3$ .

\lim_{x \to 3} (x - 4) = -1

Substitute.

Answer: define f(3) = −1.

10.

C(20) = 4 + 1.5(20) = 34

m = 20 uses the first rule.

\lim_{m \to 20^+} C(m) = 34 + 1.25(0) = 34

Right-hand limit from the second rule.

Answer: yes. The left-hand limit, the right-hand limit and C(20) are all $34, so C is continuous at 20 miles.

Continuity FAQ

What are the three conditions for continuity?

f(a) must be defined, $\lim_{x \to a} f(x)$ must exist, and the limit must equal f(a).

What are the types of discontinuity?

Removable (a hole, where the limit exists), jump (the one-sided limits differ) and infinite (a vertical asymptote).

How do you make a piecewise function continuous?

Set the left-hand and right-hand limits at the boundary equal to each other (and to the function value) and solve for the unknown constant.

Is a function with a hole continuous?

No. At the hole, f(a) is either undefined or not equal to the limit, so condition 1 or 3 fails. Filling the hole with the limit value makes it continuous.

Summary

Continuous at a: f(a) is defined, the limit exists, and the limit equals f(a).
Removable: the limit exists but is not f(a). Jump: the sides differ. Infinite: a side goes to $\pm\infty$ .
Polynomials, roots, exponentials, logs, sine and cosine are continuous on their domains; rational functions break where the denominator is 0.
For piecewise functions, compare both one-sided limits with the value at the boundary.
To find a constant that makes a function continuous, set the pieces’ values at the boundary equal.

Previous lesson: Squeeze Theorem

Next lesson: Intermediate Value Theorem

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