Continuity on an Interval: Endpoint Rules and Examples

What Is Continuity On An Interval?

Continuity on an interval means a function is continuous at every point of that interval, with the endpoints judged only from the inside. On an open interval $(a, b)$ that is the whole story. On a closed interval $[a, b]$ there are two extra checks: the function must approach $f(a)$ from the right and approach $f(b)$ from the left.

That answers one of the most-asked versions of the question, "What does it mean for a function to be continuous on an interval?" Continuity at a single point is the building block, covered in full on continuity of a function. Recall it briefly: $f$ is continuous at $c$ when $f(c)$ is defined, the limit of a function at $c$ exists, and the two agree.

$$\lim_{x \to c} f(x) = f(c)$$

Continuity on an interval stretches that idea across every point of a set. Formally, $f$ is continuous on the closed interval $[a, b]$ when all three conditions hold:

$$\text{(1)}\quad \lim_{x \to c} f(x) = f(c) \quad \text{for every } c \text{ with } a < c < b$$

$$\text{(2)}\quad \lim_{x \to a^{+}} f(x) = f(a) \quad \text{(right-continuous at } a\text{)}$$

$$\text{(3)}\quad \lim_{x \to b^{-}} f(x) = f(b) \quad \text{(left-continuous at } b\text{)}$$

Here $a$ and $b$ are the endpoints, the points strictly between them are the interior points, $x \to a^{+}$ means $x$ approaches $a$ from values larger than $a$, and $x \to b^{-}$ means $x$ approaches $b$ from values smaller than $b$. OpenStax states the same definition in Calculus Volume 1, Section 2.4. If the bracket notation itself is new, interval notation explains which brackets include an endpoint and which exclude it.

How Do Open, Closed And Half-Open Intervals Change The Check?

The interval type decides what you check at each end, and nothing else changes. Interior points always need the ordinary two-sided test. An endpoint only needs attention when the interval actually contains it.

A Quora reader asked why a function defined on something like $[1, \infty)$ is sometimes described as continuous only on $(1, \infty)$. Usually the book is being cautious, or it is using the stricter point-only definition, which needs a two-sided limit that an endpoint cannot have. Under the interval definition above, a function that is right-continuous at 1 and continuous everywhere after it is continuous on $[1, \infty)$. The two statements do not conflict: being continuous on $[1, \infty)$ automatically makes the function continuous on the smaller set $(1, \infty)$ too.

Why Does A Closed Endpoint Need Only A One-Sided Limit?

A closed endpoint needs only a one-sided limit because the interval contains nothing on the other side. At the left endpoint $a$, every point of $[a, b]$ sits to the right of $a$, so the only meaningful approach is $x \to a^{+}$. Asking for a left-hand limit there would drag in values of $x$ that the question never asked about, and often values where the function is not even defined.

The classic case is $f(x) = \sqrt{x}$ at $x = 0$. To the left of 0 the real square root does not exist, so a two-sided limit at 0 is impossible. Even so, $\sqrt{x}$ is continuous on $[0, 4]$, because $\lim_{x \to 0^{+}} \sqrt{x} = 0 = \sqrt{0}$.

Textbooks disagree in their wording here, which is part of why this trips people up: some say $\sqrt{x}$ is "continuous at 0 from the right", others say "continuous on its domain", and both mean the same thing. On a closed, bounded interval, continuity also upgrades to uniform continuity, a stronger property that real analysis courses build on.

How Do You Check Whether A Function Is Continuous On An Interval?

To check continuity on an interval, test the interior first, then each included endpoint from the inside. A Quora thread asks this almost word for word: "How to know if a function is continuous on a closed interval". These four steps answer it:

1. Find the domain. List every $x$ where the formula fails: a zero denominator, a negative number under an even root, a non-positive log argument. If any of these points lies in the interval, continuity on that interval already fails.

2. Check the interior. For standard formulas, continuity on the domain is automatic (see the next section). For a piecewise formula, test each break point inside the interval with left and right limits.

3. Check the left endpoint (if included). Confirm $\lim_{x \to a^{+}} f(x) = f(a)$.

4. Check the right endpoint (if included). Confirm $\lim_{x \to b^{-}} f(x) = f(b)$.

If all four pass, the function is continuous on the interval. If one fails, name the point and the reason, because the types of discontinuity (removable, jump, infinite) behave very differently afterward.

Which Functions Are Continuous On Every Interval Inside Their Domain?

Most functions met in school are continuous at every point of their domain, so the real work is finding the domain. A function is continuous on any interval that sits completely inside its domain, with closed endpoints allowed wherever the function is defined and approaches its value from the inside.

Rational functions are the family that catches most people, because the formula looks harmless until the denominator hits zero. A careful look at domain and range of a function settles most continuity questions before any limit is taken.

Combinations keep continuity. If $f$ and $g$ are continuous on an interval, then so are:

• Sums and differences: $f + g$ and $f - g$

• Products: $f \cdot g$

• Quotients: $\frac{f}{g}$, wherever $g(x) \neq 0$

• Compositions: $f(g(x))$, provided $f$ is continuous at the values $g$ produces

How Do You Find Intervals Of Continuity From A Graph?

On a graph, the intervals of continuity are the stretches you can trace without lifting your pencil, with the dots at each end deciding whether the endpoint is included. "What are the regions of continuity for the following function?" is one of the most common homework Q&A titles on the topic, and it is nearly always a graph-reading question.

Read the graph left to right and stop at every break:

• A hole (hollow dot with no filled dot above or below it) splits the graph and is not included.

• A jump splits the graph. The side with the filled dot gets the endpoint. For example, if a jump discontinuity at $x = 2$ has its filled dot on the left piece, write $[0, 2]$ and $(2, 5]$.

• A vertical asymptote splits the graph, and neither side includes it.

The pencil test is a good first read. Actually, "test" oversells it: a graph cannot prove continuity at a single point. Use the graph to locate the suspicious points, then confirm each one with limits.

Examples Of Continuity On An Interval

Example 1

Is $f(x) = \sqrt{9 - x^2}$ continuous on $[-3, 3]$?

The domain is $9 - x^2 \geq 0$, which is exactly $-3 \leq x \leq 3$.

Interior: a root of a polynomial is continuous wherever the radicand is positive, so every $c$ in $(-3, 3)$ passes.

Left endpoint:

$$\lim_{x \to -3^{+}} \sqrt{9 - x^2} = 0 = f(-3)$$

Right endpoint:

$$\lim_{x \to 3^{-}} \sqrt{9 - x^2} = 0 = f(3)$$

Final answer: Yes. $f$ is continuous on $[-3, 3]$ (its graph is the upper half of a circle of radius 3).

Example 2

Is $f(x) = \sqrt{x}$ continuous on $[0, 4]$?

The tempting route is to test $x = 0$ the way you would test any point, with a two-sided limit.

$$\lim_{x \to 0^{-}} \sqrt{x} \quad \text{does not exist (no real square roots of negatives)}$$

So the two-sided limit at 0 does not exist, and the conclusion seems to be "not continuous at 0, so not continuous on $[0, 4]$." That cannot be right: the graph of $\sqrt{x}$ starts at the origin and rises without a single gap. The error is the left-hand limit, which asks about $x < 0$, a region outside the interval.

The correct check uses only the inside of the interval:

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

$$\lim_{x \to 4^{-}} \sqrt{x} = 2 = f(4)$$

Every interior point passes because $\sqrt{x}$ is continuous for $x > 0$.

Final answer: Yes. $\sqrt{x}$ is continuous on $[0, 4]$.

Example 3

On which of $[0, 2)$ and $[0, 3]$ is $g(x) = \frac{1}{x - 2}$ continuous?

The denominator is zero at $x = 2$.

On $[0, 2)$, the point 2 is excluded. At the left endpoint:

$$\lim_{x \to 0^{+}} \frac{1}{x - 2} = -\frac{1}{2} = g(0)$$

So $g$ is continuous on $[0, 2)$. The fact that $g(x) \to -\infty$ as $x \to 2^{-}$ does not matter, because 2 is not in the interval.

On $[0, 3]$, the point $x = 2$ is inside and $g(2)$ is undefined.

Final answer: Continuous on $[0, 2)$, not continuous on $[0, 3]$.

The first instinct on this problem is to check only the two endpoints, 0 and 3, find both values defined, and declare the function continuous. The break sits in the middle, which is why step 1 of the checklist is always the domain.

Example 4

Is $h(x) = \ln(x - 1)$ continuous on $(1, 5]$? On $[1, 5]$?

The log needs $x - 1 > 0$, so the domain is $x > 1$.

On $(1, 5]$: the left end is excluded, and at the right end

$$\lim_{x \to 5^{-}} \ln(x - 1) = \ln 4 \approx 1.3863 = h(5)$$

On $[1, 5]$: $h(1) = \ln 0$ is undefined, and

$$\lim_{x \to 1^{+}} \ln(x - 1) = -\infty$$

Final answer: Continuous on $(1, 5]$, not continuous on $[1, 5]$.

Example 5

Find $k$ so that $f$ is continuous on $[0, 4]$, where $f(x) = x^2 + k$ for $0 \leq x < 2$ and $f(x) = 3x - 1$ for $2 \leq x \leq 4$.

Each piece is a polynomial, so the only interior point in doubt is the break at $x = 2$. The endpoints 0 and 4 each sit inside a single polynomial piece, so they pass automatically.

Left-hand limit at 2:

$$\lim_{x \to 2^{-}} (x^2 + k) = 4 + k$$

Value (and right-hand limit) at 2:

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

Set them equal:

$$4 + k = 5$$

$$k = 1$$

Final answer: $k = 1$. (For more problems that match pieces at a break point, see piecewise functions.)

Example 6

Can the Intermediate Value Theorem show that $f(x) = \frac{x + 1}{(x - 3)(x + 2)}$ has a zero on $[-3, 4]$? On $\left[-\frac{3}{2}, 2\right]$?

On $[-3, 4]$ the values at the ends are

$$f(-3) = \frac{-2}{(-6)(-1)} = -\frac{1}{3}$$

$$f(4) = \frac{5}{(1)(6)} = \frac{5}{6}$$

The signs differ, but $x = -2$ and $x = 3$ both lie inside $[-3, 4]$, where $f$ is undefined. The theorem's hypothesis fails, so it says nothing on this interval.

On $\left[-\frac{3}{2}, 2\right]$, neither $-2$ nor $3$ is inside, so $f$ is continuous there. The end values are

$$f\left(-\tfrac{3}{2}\right) = \frac{-\frac{1}{2}}{\left(-\frac{9}{2}\right)\left(\frac{1}{2}\right)} = \frac{2}{9}$$

$$f(2) = \frac{3}{(-1)(4)} = -\frac{3}{4}$$

The signs differ and $f$ is continuous on the interval, so a zero exists between $-\frac{3}{2}$ and 2. Solving $x + 1 = 0$ confirms it is $x = -1$.

Final answer: Not on $[-3, 4]$ (hypothesis fails). Yes on $\left[-\frac{3}{2}, 2\right]$, where the zero is $x = -1$.

Why Does Continuity On An Interval Matter?

Continuity on an interval matters because three of the most useful theorems in calculus guarantee their results only on a closed interval with no breaks. Each one says "something must happen somewhere in here", and each one fails the moment a single point breaks. This is the honest answer to the Quora question "Is it important to understand continuity of intervals in calculus?"

• The Intermediate Value Theorem. If $f$ is continuous on $[a, b]$, it takes every value between $f(a)$ and $f(b)$. This is how you prove that $x^3 + x - 1 = 0$ has a solution between 0 and 1: $f(0) = -1$, $f(1) = 1$, and the root sits near $x \approx 0.6823$.

• The Extreme Value Theorem. If $f$ is continuous on $[a, b]$, it reaches an absolute maximum and an absolute minimum there. Drop the closed interval and the promise disappears: $\frac{1}{x}$ on $(0, 1]$ grows without bound, and $f(x) = x$ on $(0, 1)$ has neither a largest nor a smallest value (see the extreme value theorem entry for the same pair). On $\left[\frac{1}{2}, 1\right]$, the same $\frac{1}{x}$ has maximum 2 and minimum 1.

• The Mean Value Theorem. It needs continuity on $[a, b]$ and differentiability on $(a, b)$.

That last pairing answers another Quora favourite: "Why do we talk about continuity of a real valued function in a closed interval and differentiability of the same function in an open interval?" Continuity at an endpoint needs only a one-sided limit, which is cheap. The theorem needs the endpoint values $f(a)$ and $f(b)$ to be tied to the curve, so it asks for continuity there, but it never needs a slope at the endpoints, so it asks for differentiability only inside.

The idea has a real origin story. In 1817 the Prague mathematician Bernard Bolzano published a "purely analytic proof" of the intermediate value property. Before that, the fact that a continuous curve crossing an axis must hit it was treated as geometrically obvious. Bolzano's insistence on a proof is a large part of why continuity is now defined with limits instead of with pictures.

Where Is Continuity On An Interval Used In Real Life?

Continuity on an interval is the hidden assumption behind any method that searches for a value between two known values. A few places it does real work:

• Root-finding software. The bisection method keeps halving an interval where the function changes sign, which is only safe when the function is continuous on that interval.

• Measured quantities. Temperature, altitude and position change without teleporting, so if a reading goes from 18 degrees to 26 degrees between morning and afternoon, it passed through 22 degrees at some moment.

• Optimisation on a fixed range. When a cost or profit model is continuous on a closed production range, the Extreme Value Theorem guarantees a best and a worst output level exist before you start hunting for them with derivatives.

Where Do Students Trip Up On Continuity On An Interval?

Most lost marks on this topic come from the endpoints, not the interior. The three mistakes below are the ones that show up again and again.

Mistake 1: Demanding A Two-Sided Limit At A Closed Endpoint

Where it slips in:

Checking $\sqrt{x}$ on $[0, 4]$, or any function whose domain starts exactly at the interval's left end.

Don't do this:

Do not test $\lim_{x \to 0} \sqrt{x}$ as a two-sided limit and conclude "discontinuous at 0".

The correct way:

At a left endpoint use only $x \to a^{+}$; at a right endpoint use only $x \to b^{-}$.

The rusher makes this mistake most often: they apply the three-part point test on autopilot and never ask which side of the endpoint the interval actually lives on.

Mistake 2: Swapping Which One-Sided Limit Goes With Which Endpoint

Where it slips in:

Writing the closed-interval definition from memory under time pressure.

Don't do this:

Do not write "left-continuous at $a$, right-continuous at $b$". The left endpoint is approached from the right, and the right endpoint from the left.

The correct way:

Picture standing inside the interval and walking toward each end. At $a$ you arrive from the right ($x \to a^{+}$), and at $b$ you arrive from the left ($x \to b^{-}$).

The memorizer is most exposed here. They can recite "left and right" but cannot rebuild which goes where, so the fix is the walking picture, not a better mnemonic. Students who sketch a two-arrow number line before writing any limit almost never swap them.

Mistake 3: Checking Only The End Values Before Using A Theorem

Where it slips in:

Intermediate Value Theorem problems, where $f(a)$ and $f(b)$ have opposite signs and the conclusion feels automatic.

Don't do this:

Do not claim a root of $\tan x$ in $[1, 2]$ just because $\tan 1 > 0$ and $\tan 2 < 0$.

The correct way:

Confirm continuity on the whole interval first. Here $\tan x$ breaks at $x = \frac{\pi}{2} \approx 1.5708$, so the theorem does not apply, and in fact there is no root.

Here is what this mistake looks like outside a classroom. The bisection method requires a continuous function with opposite signs at the two ends. Feed it $\tan x$ on $[1, 2]$ anyway, and after 40 halvings it settles on $x \approx 1.5707963$, which is $\frac{\pi}{2}$, the asymptote and not a root.

At that point $\tan x$ is about $-6 \times 10^{12}$, nowhere near zero. A program that skips the continuity check can report that point as a "root" with full confidence.

Practice Problems On Continuity On An Interval

1. Is $f(x) = \sqrt{x - 3}$ continuous on $[3, 7]$?

2. Is $g(x) = \frac{1}{x^2 - 1}$ continuous on $[0, 1]$? On $[0, 1)$?

3. Find $k$ so that $h(x) = kx + 1$ for $x < 1$ and $h(x) = x^2 + 2$ for $x \geq 1$ is continuous on $[-2, 3]$.

4. Use continuity on an interval to show that $x^3 - 3x + 1 = 0$ has a root in $(0, 1)$.

5. On which interval is $f(x) = \ln(4 - x^2)$ continuous? Is it continuous on $[-2, 2]$?

Answer to Problem 1: Yes. $\lim_{x \to 3^{+}} \sqrt{x - 3} = 0 = f(3)$, the interior is fine, and $f(7) = 2$ matches the left-hand limit.

Answer to Problem 2: Not on $[0, 1]$, because $g(1)$ is undefined and $g(x) \to -\infty$ as $x \to 1^{-}$. Yes on $[0, 1)$, with $g(0) = -1$.

Answer to Problem 3: $k = 2$, since $k + 1 = 1^2 + 2 = 3$.

Answer to Problem 4: The polynomial is continuous on $[0, 1]$, $f(0) = 1$ and $f(1) = -1$, so a root exists (it is near $x \approx 0.3473$).

Answer to Problem 5: The domain is $-2 < x < 2$, so $f$ is continuous on $(-2, 2)$ but not on $[-2, 2]$, because $\ln 0$ is undefined.

Wrapping Up

• Continuity on an interval means continuity at every interior point plus one-sided continuity at each endpoint the interval includes.

• At a left endpoint the check is $x \to a^{+}$, and at a right endpoint it is $x \to b^{-}$, because the interval has nothing on the far side.

• The domain decides most cases: polynomials, roots, logs, trig and rational functions are continuous on any interval inside their domains.

• The Intermediate Value Theorem and Extreme Value Theorem need continuity on a closed interval, and a single break makes their guarantee disappear.

• The most costly mistake is checking only the end values, as the $\tan x$ example on $[1, 2]$ shows.

Your Next Move

Work through practice problems 1 to 5 in order; if Problem 2 or Problem 5 trips you up, return to the section on open, closed and half-open intervals and the endpoint table.

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Read More

• Continuity and differentiability

• Differentiability vs continuity

• Removable discontinuity

• Infinite discontinuity

• Rolle's theorem

• Limit laws