Continuity

Informally, a function is continuous if you can draw its graph without lifting your pencil: no holes, no jumps, no asymptotes. Calculus needs a precise version of this idea, because big theorems like the Intermediate Value Theorem only work for continuous functions. The definition uses limits, and on the AP exam you’ll need to check it step by step.

Key ideas

Continuity at a point

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

1. $f(a)$ is defined.
2. $\displaystyle\lim_{x \to a} f(x)$ exists.
3. $\displaystyle\lim_{x \to a} f(x) = f(a)$.

If any condition fails, $f$ is discontinuous at $a$. Which condition fails tells you what kind of break it is (see types of discontinuities).

Condition 3 is the heart of it: the value the function approaches is the value it has. That’s also why, for continuous functions, you can find limits by substituting.

The graph of $f$ for Example 1: check continuity at $x = -2$, $0$, $2$, and $4$.

Continuity on an interval

• $f$ is continuous on an open interval $(a, b)$ if it is continuous at every point inside.
• $f$ is continuous on a closed interval $[a, b]$ if it is continuous on $(a, b)$, and at the endpoints the one-sided limits match: $\displaystyle\lim_{x \to a^+} f(x) = f(a)$ and $\displaystyle\lim_{x \to b^-} f(x) = f(b)$.

Functions that are continuous on their domains

These families are continuous at every point of their domain:

Function type	Where it’s continuous
Polynomials	all real numbers
Rational functions $\dfrac{p(x)}{q(x)}$	everywhere except where $q(x) = 0$
$\sqrt{x}$ (even roots)	$x \ge 0$
$\sqrt[3]{x}$ (odd roots)	all real numbers
$e^x$, $b^x$	all real numbers
$\ln x$	$x \gt 0$
$\sin x$, $\cos x$	all real numbers
$\tan x$	everywhere except $x = \dfrac{\pi}{2} + k\pi$ (where $\cos x = 0$)

Sums, differences, products, and compositions of continuous functions are continuous, and so are quotients wherever the denominator isn’t $0$. So to find where a formula is continuous, find its domain.

Piecewise functions are the exception: each piece may be continuous, but you have to check the boundary points with the three conditions.

AP-style justification

To show $f$ is continuous at $a$, show all three: ”$f(2) = 3$. $\displaystyle\lim_{x \to 2^-} f(x) = 3$ and $\displaystyle\lim_{x \to 2^+} f(x) = 3$, so $\displaystyle\lim_{x \to 2} f(x) = 3$. Since $\displaystyle\lim_{x \to 2} f(x) = f(2)$, $f$ is continuous at $x = 2$.”

Worked examples

Example 1: Reading continuity from a graph

Use the graph of $f$ above. At each of $x = -2$, $0$, $2$, and $4$, decide whether $f$ is continuous. If not, say which condition fails.

Solution.

• $x = -2$: $f(-2) = 3$ (filled dot), and $\displaystyle\lim_{x \to -2} f(x) = 1$ (the hole). The limit exists but doesn’t equal $f(-2)$. Not continuous: condition 3 fails.
• $x = 0$: the curve passes smoothly through $(0, 2)$. $f(0) = 2 = \displaystyle\lim_{x \to 0} f(x)$. Continuous.
• $x = 2$: the left-hand limit is $5$ and the right-hand limit is $1$, so the limit doesn’t exist. Not continuous: condition 2 fails.
• $x = 4$: there’s a hole and no dot, so $f(4)$ is undefined. Not continuous: condition 1 fails. (The limit does exist; it’s $3$.)

Example 2: A piecewise function

Is $g(x) = \begin{cases} x^2 - 1, & x \lt 2 \\ 2x - 1, & x \ge 2 \end{cases}$ continuous at $x = 2$? Justify your answer.

Solution. Check all three conditions.

1. $g(2) = 2(2) - 1 = 3$, so $g(2)$ is defined.
2. $\displaystyle\lim_{x \to 2^-} (x^2 - 1) = 3$ and $\displaystyle\lim_{x \to 2^+} (2x - 1) = 3$. Both sides agree, so $\displaystyle\lim_{x \to 2} g(x) = 3$.
3. $\displaystyle\lim_{x \to 2} g(x) = 3 = g(2)$.

All three conditions hold, so $g$ is continuous at $x = 2$.

Example 3: Where is a function continuous?

Where is $f(x) = \dfrac{x + 1}{x^2 - 9}$ continuous? Where is $h(x) = \sqrt{x - 2}$ continuous?

Solution. $f$ is a rational function, so it’s continuous everywhere except where the denominator is $0$: $x^2 - 9 = 0$ at $x = \pm 3$. So $f$ is continuous on $\{x \in \mathbb{R} \mid x \ne \pm 3\}$.

$h$ is a square root, continuous wherever $x - 2 \ge 0$. So $h$ is continuous on $\{x \in \mathbb{R} \mid x \ge 2\}$. At $x = 2$ it’s continuous from the right only, which is all we need at an endpoint.

Example 4: Combining functions

Where is $k(x) = \ln(x - 1) + \sqrt{5 - x}$ continuous?

Solution. Each piece is continuous on its domain:

• $\ln(x - 1)$ needs $x - 1 \gt 0$, so $x \gt 1$.
• $\sqrt{5 - x}$ needs $5 - x \ge 0$, so $x \le 5$.

The sum is continuous where both parts are: $\{x \in \mathbb{R} \mid 1 \lt x \le 5\}$.

Common mistakes

Checking only one or two conditions. Showing the limit exists is not enough; you must also show it equals $f(a)$. On the AP exam, a continuity justification needs all three conditions, with numbers.

Assuming “defined” means “continuous”. At $x = -2$ in Example 1, $f(-2)$ is defined, yet $f$ is not continuous there.

Forgetting the boundary points of piecewise functions. Each piece of $g$ in Example 2 is a polynomial, but that doesn’t guarantee continuity at $x = 2$. Check the boundary every time.

Mixing up domain and continuity for rational functions. A function like $\dfrac{x^2 - 9}{x - 3}$ is not continuous at $x = 3$, even though it simplifies to $x + 3$. It isn’t defined there.

Using degrees for trig. Write the discontinuities of $\tan x$ as $x = \dfrac{\pi}{2} + k\pi$ in radians, not $90^\circ + 180^\circ k$.

Practice

1. (Warm-up) Suppose $f(3) = 4$ and $\displaystyle\lim_{x \to 3} f(x) = 4$. Is $f$ continuous at $x = 3$?

Solution

Yes. $f(3)$ is defined, the limit exists, and they are equal, so all three conditions hold.

2. (Warm-up) Suppose $\displaystyle\lim_{x \to 1} f(x) = 2$ and $f(1) = 5$. Is $f$ continuous at $x = 1$? If not, which condition fails?

Solution

No. $f(1)$ is defined and the limit exists, but $2 \ne 5$, so condition 3 fails.

3. (Warm-up) Where is $p(x) = x^3 - 4x + 1$ continuous?

Solution

$p$ is a polynomial, so it’s continuous for all real numbers.

4. (Core) Where is $r(x) = \dfrac{x - 2}{x^2 - 5x + 6}$ continuous?

Solution

The denominator factors as $(x - 2)(x - 3)$, which is $0$ at $x = 2$ and $x = 3$. So $r$ is continuous on $\{x \in \mathbb{R} \mid x \ne 2, 3\}$.

(Even though the factor $x - 2$ cancels, $r(2)$ is undefined, so $r$ is not continuous at $2$.)

5. (Core) Is $g(x) = \begin{cases} 3x + 1, & x \le 1 \\ x^2 + 3, & x \gt 1 \end{cases}$ continuous at $x = 1$? Justify your answer.

Solution

1. $g(1) = 3(1) + 1 = 4$.
2. $\displaystyle\lim_{x \to 1^-} (3x + 1) = 4$ and $\displaystyle\lim_{x \to 1^+} (x^2 + 3) = 4$, so $\displaystyle\lim_{x \to 1} g(x) = 4$.
3. $\displaystyle\lim_{x \to 1} g(x) = g(1)$.

So $g$ is continuous at $x = 1$.

6. (Core) Is $h(x) = \begin{cases} x^2, & x \lt 0 \\ \cos x, & x \ge 0 \end{cases}$ continuous at $x = 0$? (Radians.)

Solution

$\displaystyle\lim_{x \to 0^-} x^2 = 0$ and $\displaystyle\lim_{x \to 0^+} \cos x = \cos 0 = 1$. The one-sided limits differ, so $\displaystyle\lim_{x \to 0} h(x)$ does not exist. $h$ is not continuous at $0$ (condition 2 fails).

7. (Core) Where is $f(x) = \dfrac{\sqrt{x + 3}}{x - 1}$ continuous?

Solution

The square root needs $x + 3 \ge 0$, so $x \ge -3$. The denominator needs $x \ne 1$.

So $f$ is continuous on $\{x \in \mathbb{R} \mid x \ge -3,\ x \ne 1\}$.

8. (Challenge) Show that $f(x) = \begin{cases} x^2, & x \le -1 \\ 2x + 3, & -1 \lt x \lt 2 \\ 9 - x, & x \ge 2 \end{cases}$ is continuous for all real numbers.

Solution

Each piece is a polynomial, so $f$ is continuous everywhere except possibly at the boundaries $x = -1$ and $x = 2$.

At $x = -1$: $f(-1) = (-1)^2 = 1$. $\displaystyle\lim_{x \to -1^-} x^2 = 1$ and $\displaystyle\lim_{x \to -1^+} (2x + 3) = 1$. The limit is $1 = f(-1)$, so $f$ is continuous at $-1$.

At $x = 2$: $f(2) = 9 - 2 = 7$. $\displaystyle\lim_{x \to 2^-} (2x + 3) = 7$ and $\displaystyle\lim_{x \to 2^+} (9 - x) = 7$. The limit is $7 = f(2)$, so $f$ is continuous at $2$.

So $f$ is continuous for all real numbers.

9. (Challenge) Let $h(x) = \begin{cases} x\sin\left(\dfrac{1}{x}\right), & x \ne 0 \\ 0, & x = 0 \end{cases}$. Show that $h$ is continuous at $x = 0$.

Solution

$h(0) = 0$ is defined. For the limit, use the squeeze theorem: since $\left|\sin\left(\dfrac{1}{x}\right)\right| \le 1$,

$$-|x| \le x\sin\left(\frac{1}{x}\right) \le |x|$$

Both bounds approach $0$ as $x \to 0$, so $\displaystyle\lim_{x \to 0} h(x) = 0$.

Since $\displaystyle\lim_{x \to 0} h(x) = 0 = h(0)$, $h$ is continuous at $0$.