Defining Continuity at a Point (AP Calculus AB 1.11)

Drawing a graph without lifting your pencil is a good picture of continuity, but a picture can't settle whether $f(x) = \frac{\sin 3x}{x}$ , with $f(0)$ set equal to 3, is continuous at 0. For that you need a definition that turns "no break in the graph" into statements about limits you can check one at a time.

The three conditions

At $x = c$ , a graph has no break exactly when the curve arrives at the same height from both sides and the point $(c, f(c))$ sits at that height. Written as conditions:

The third condition is the heart of it, and it can't even be stated unless the first two hold, which is why they are listed separately: checking them in order tells you exactly what went wrong. Each failure matches a type of discontinuity from topic 1.10.

Condition that fails first	What the graph shows	Type
1: $f(c)$ undefined	no point on the line $x = c$ : a hole, a gap between two pieces, or an asymptote	removable if the limit exists; otherwise jump or infinite, or neither named type for an oscillation like $\sin\left(\frac{1}{x}\right)$ at 0 (topic 1.10)
2: no finite two-sided limit	the curve arrives at different heights, runs off to $\pm\infty$ , or oscillates without settling	jump or infinite, or neither named type for an oscillation (such as $\sin\left(\frac{1}{x}\right)$ given a value at 0)
3: limit $\neq f(c)$	a hole with the point drawn somewhere else	removable

Here is the definition settling the question from the opening. Let $f(x) = \frac{\sin 3x}{x}$ for $x \neq 0$ and $f(0) = 3$ .

$f(0) = 3$ , so the value exists.
Write $\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}$ . As $x \to 0$ , $3x \to 0$ , and $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ (topic 1.8), so the limit is $3 \cdot 1 = 3$ .
The limit, 3, equals $f(0)$ .

All three hold, so $f$ is continuous at 0. Had the definition said $f(0) = 1$ , conditions 1 and 2 would still hold and condition 3 would fail.

Continuity is what makes substitution legal

Condition 3 read backward says: if you know $f$ is continuous at $c$ , then $\lim\limits_{x \to c} f(x)$ is simply $f(c)$ . That is the real reason direct substitution worked in topics 1.5 and 1.6. Polynomials and the other basic functions are continuous wherever they are defined (topic 1.12 lists them), so their limits are their values. Substitution fails exactly at points where a function is not continuous, which is why $\frac{x^2 - 9}{x - 3}$ at $x = 3$ needed factoring.

The same reading works with information you are handed. If you are told $f$ is continuous at 2 and $f(2) = 5$ , then $\lim\limits_{x \to 2} f(x) = 5$ , and limit laws give, say, $\lim\limits_{x \to 2}\,[\,3f(x) - x^2\,] = 15 - 4 = 11$ , without any formula for $f$ . This is how free-response questions with tables use continuity: the stated continuity turns a table value into a limit.

Writing the justification

A free-response question that asks "Is $f$ continuous at $x = c$ ? Justify your answer" awards its point for showing the conditions, not for the word "yes" or "no." A complete justification for a piecewise function has this shape:

the value $f(c)$ , computed from the piece that includes $x = c$ ;
the left-hand limit, from the piece used for $x < c$ ;
the right-hand limit, from the piece used for $x > c$ ;
the conclusion, tied to the definition: "Since $\lim\limits_{x \to c} f(x) = f(c)$ , $f$ is continuous at $x = c$ ," or the specific condition that fails.

For a single formula that is not defined at $c$ , the shortest correct justification is condition 1: " $f(c)$ is undefined, so $f$ is not continuous at $c$ ." For a function defined at $c$ by a special value, you compute the limit from the general formula and compare it with the special value, as in the opening example.

Worked examples

Worked exampleExample 1: a piecewise function at its boundary

Let

h(x) = \begin{cases} x^3 - 2x & x \le 2 \\ 12 - 2x^2 & x > 2 \end{cases}

Is $h$ continuous at $x = 2$ ? Justify your answer.

Show solution

Check the three conditions, using the piece that applies on each side.

The point $x = 2$ belongs to the first piece, so $h(2) = 8 - 4 = 4$ .
For $x < 2$ , $h(x) = x^3 - 2x$ , a polynomial, so $\lim\limits_{x \to 2^-} h(x) = 8 - 4 = 4$ . For $x > 2$ , $h(x) = 12 - 2x^2$ , so $\lim\limits_{x \to 2^+} h(x) = 12 - 8 = 4$ . The one-sided limits agree, so $\lim\limits_{x \to 2} h(x) = 4$ .
$\lim\limits_{x \to 2} h(x) = 4 = h(2)$ .

Since $\lim\limits_{x \to 2} h(x) = h(2)$ , $h$ is continuous at $x = 2$ . Each one-sided limit was found by substitution because each piece is a polynomial, and polynomials are continuous.

Worked exampleExample 2: a special value that doesn't match

Let $g(x) = \dfrac{\sqrt{x + 9} - 3}{x}$ for $x \neq 0$ , and $g(0) = \dfrac{1}{3}$ . Is $g$ continuous at $x = 0$ ?

Show solution

Condition 1 holds: $g(0) = \frac{1}{3}$ .

For condition 2, substitution into the formula gives $\frac{0}{0}$ , so rewrite. Multiply the numerator and the denominator by the conjugate $\sqrt{x + 9} + 3$ :

\frac{\sqrt{x + 9} - 3}{x}\cdot\frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} = \frac{(x + 9) - 9}{x\left(\sqrt{x + 9} + 3\right)} = \frac{1}{\sqrt{x + 9} + 3} \quad (x \neq 0).

The last expression is continuous at 0, so $\lim\limits_{x \to 0} g(x) = \dfrac{1}{3 + 3} = \dfrac{1}{6}$ . The limit exists.

Condition 3 fails: $\lim\limits_{x \to 0} g(x) = \frac{1}{6} \neq \frac{1}{3} = g(0)$ . So $g$ is not continuous at $x = 0$ . The discontinuity is removable: redefining $g(0) = \frac{1}{6}$ would fix it.

Worked exampleExample 3: an exam-style free-response question (no calculator)

Let

f(x) = \begin{cases} 1 + e^{x - 1} & x < 1 \\ 2 + \ln x & 1 \le x \le e \\ 5 - \dfrac{x}{e} & x > e \end{cases}

(a) Is $f$ continuous at $x = 1$ ? Justify your answer.

(b) Is $f$ continuous at $x = e$ ? Justify your answer.

Show solution

(a) $f(1) = 2 + \ln 1 = 2$ . $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-}\left(1 + e^{x - 1}\right) = 1 + e^0 = 2$ . $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+}\left(2 + \ln x\right) = 2 + 0 = 2$ . So $\lim\limits_{x \to 1} f(x) = 2 = f(1)$ , and $f$ is continuous at $x = 1$ .

(b) $f(e) = 2 + \ln e = 3$ . $\lim\limits_{x \to e^-} f(x) = \lim\limits_{x \to e^-}\left(2 + \ln x\right) = 3$ . $\lim\limits_{x \to e^+} f(x) = \lim\limits_{x \to e^+}\left(5 - \dfrac{x}{e}\right) = 5 - 1 = 4$ . The one-sided limits are not equal, so $\lim\limits_{x \to e} f(x)$ does not exist, and $f$ is not continuous at $x = e$ . (Condition 2 fails; it is a jump.)

(c) $k(e) = (e - e)f(e) = 0 \cdot 3 = 0$ . $\lim\limits_{x \to e^-} k(x) = 0 \cdot 3 = 0$ and $\lim\limits_{x \to e^+} k(x) = 0 \cdot 4 = 0$ , since $x - e \to 0$ from both sides while $f$ stays near 3 or 4. So $\lim\limits_{x \to e} k(x) = 0 = k(e)$ , and $k$ is continuous at $x = e$ .

Where students lose points

On the exam

"Is $f$ continuous at $x = c$ ? Justify your answer" can be a free-response part. The point goes to work that shows the value and both one-sided limits (for a piecewise function, each from the piece used on that side) and states that the limit equals the value. A question about differentiability at a point (topic 2.4) needs this check too, since a function must be continuous at a point to be differentiable there. Multiple-choice versions give a piecewise rule or a graph and ask at which points the function is continuous, or which condition fails.

Check yourself

Question 1
Let $f(x) = \dfrac{x^2 - 16}{x - 4}$ for $x \neq 4$ and $f(4) = 4$ . Which statement describes $f$ at $x = 4$ ?
1. A
  $f(4)$ is undefined, so $f$ is not continuous at 4
2. B
  $\lim\limits_{x \to 4} f(x)$ does not exist, so $f$ is not continuous at 4
3. C
  $\lim\limits_{x \to 4} f(x)$ exists but does not equal $f(4)$ , so $f$ is not continuous at 4
4. D
  $f$ is continuous at 4
Show answer
C
For $x \neq 4$ , $f(x) = x + 4$ , so the limit is 8, while $f(4) = 4$ . Conditions 1 and 2 hold and condition 3 fails. Choice B is what you get if you stop at the $\frac{0}{0}$ from substitution instead of simplifying.
Question 2
The function $f$ is continuous at $x = 2$ , and $f(2) = 5$ . Find $\lim\limits_{x \to 2}\left([f(x)]^2 - x\right)$ .
Show answer
$23$
Continuity at 2 means $\lim\limits_{x \to 2} f(x) = f(2) = 5$ . Then the limit laws give $5^2 - 2 = 23$ . Nothing else about $f$ is needed.
Question 3
Let $g(x) = x^2 + 1$ for $x < 0$ and $g(x) = \cos x$ for $x \ge 0$ . Is $g$ continuous at $x = 0$ ? Give the three quantities that decide it.
Show answer
Yes. $g(0) = \cos 0 = 1$ , $\lim\limits_{x \to 0^-} g(x) = 0 + 1 = 1$ , and $\lim\limits_{x \to 0^+} g(x) = \cos 0 = 1$ , so $\lim\limits_{x \to 0} g(x) = 1 = g(0)$ .
The value comes from the piece whose inequality includes 0, $\cos x$ . Each one-sided limit comes from the piece used on that side, and both pieces are continuous, so substitution gives them. All three numbers agree, so all three conditions hold.
Question 4
Which of the following, by itself, guarantees that $f$ is continuous at $x = 3$ ?
1. A
  $\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^+} f(x)$
2. B
  $f(3) = 7$ and $f$ is defined for all real numbers
3. C
  $\lim\limits_{x \to 3^-} f(x) = f(3)$
4. D
  $\lim\limits_{x \to 3} f(x) = 7$ and $f(3) = 7$
Show answer
D
D states conditions 2 and 3 directly, and condition 1 is included because $f(3) = 7$ is given. A allows a removable discontinuity (equal one-sided limits, wrong or missing value). B says nothing about limits. C checks only the left side, so a jump on the right is still possible.

Go deeper

OpenStax, Calculus Volume 1, 2.4 Continuity