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 a test 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.

Writing a full 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 $f(a)$. A full continuity justification (on the AP exam, IB, or a class test) 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 $0$.