In short

A function is continuous at a point if three things are true: the function is defined there, the limit exists there, and the limit equals the function's value. Geometrically, this means the graph has no hole, no jump, and no vertical explosion at that point. A function that is continuous at every point in an interval can be drawn in one unbroken stroke across that interval.

Take the function f(x) = x^2 and plot it. You get a smooth parabola — a single, unbroken curve that you can trace with your finger from left to right without ever lifting your hand off the screen. The graph has no gaps, no jumps, no sudden teleportations.

Now take this function:

g(x) = \begin{cases} x^2 & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases}

This looks almost identical to f(x) = x^2, except at the single point x = 2. At that point, the parabola would give you 4, but g has been tampered with — someone reached in and moved the value at x = 2 up to 5. The graph has a hole at (2, 4) and a lonely dot floating at (2, 5).

The function $g(x)$: identical to $x^2$ everywhere except at $x = 2$, where the value has been shifted from $4$ to $5$. The open circle at $(2, 4)$ marks the hole; the red dot at $(2, 5)$ marks the actual value. Your pen would have to jump.

If you tried to draw g without lifting your pen, you would be fine everywhere — until you reached x = 2. There, you would have to lift off the curve at the hole and stamp a dot at (2, 5). The graph is broken at that point.

The word for "not broken" is continuous. A function is continuous at a point if the graph passes through that point smoothly — no holes, no jumps, no explosions. The function f(x) = x^2 is continuous at x = 2. The function g(x) is not.

That is the intuition. But "no holes, no jumps, no explosions" is not a definition you can work with in mathematics. You need something precise enough to prove things with. The precise version is built from limits.

What continuity really means

Here is what goes wrong at x = 2 for the function g. As x approaches 2 — from either side — the values g(x) approach 4, because near x = 2 the function is just x^2. The limit is 4. But the actual value g(2) is 5. The limit and the value disagree.

That disagreement is the entire problem. For a function to be continuous at a point, the limit has to match the value. No disagreement. No surprise when you arrive.

Think of it like walking toward a door. As you get closer, you can see through the glass that the room inside looks a certain way. Continuity means that when you actually step through the door, the room is exactly what you saw through the glass. Discontinuity means the room suddenly looks different once you are inside — something changed at the threshold.

Three things can go wrong, and any one of them breaks continuity:

  1. The function is not defined at the point. There is no room — the door leads to a wall.
  2. The limit does not exist at the point. You cannot tell what the room looks like through the glass — different windows show different views.
  3. The limit exists, but it does not equal the function's value. You can see through the glass just fine, but the room changes the instant you step inside.

Continuity at a point is the demand that none of these three things go wrong.

Continuity at a point

A function f is continuous at x = a if 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 one of these fails, f is discontinuous at x = a.

Condition 3 is the one that does the heavy lifting. It says: the value the function is approaching as you get close to a must be the same as the value the function actually takes at a. No surprises. No teleportation.

Notice that condition 3 implicitly requires conditions 1 and 2. If f(a) is not defined, you cannot write \lim_{x \to a} f(x) = f(a) — the right side is meaningless. If the limit does not exist, you cannot write the equality either. So some textbooks compress all three into the single statement \lim_{x \to a} f(x) = f(a). That is correct, but listing the three conditions separately is clearer when you are checking them one at a time.

Left and right continuity

Limits can be one-sided. A function can approach one value from the left and a different value from the right. When continuity depends on which side you approach from, you get one-sided continuity.

Left and right continuity

  • f is left-continuous (or continuous from the left) at x = a if \displaystyle\lim_{x \to a^-} f(x) = f(a).
  • f is right-continuous (or continuous from the right) at x = a if \displaystyle\lim_{x \to a^+} f(x) = f(a).

f is continuous at x = a if and only if it is both left-continuous and right-continuous there.

When would you care about this distinction? When the function behaves differently on the two sides of a point. The classic case is the greatest integer function f(x) = \lfloor x \rfloor — the function that takes any real number and rounds it down to the nearest integer.

Take x = 3. From the right, as x comes down from values like 3.1, 3.01, 3.001, the floor stays at 3. So \lim_{x \to 3^+} \lfloor x \rfloor = 3, which equals f(3) = 3. Right-continuous: yes.

From the left, as x approaches 3 through values like 2.9, 2.99, 2.999, the floor is 2 — always 2, right up until the instant x reaches 3. So \lim_{x \to 3^-} \lfloor x \rfloor = 2, which does not equal f(3) = 3. Left-continuous: no.

The greatest integer function $\lfloor x \rfloor$. At each integer, the function jumps up by $1$. The filled dots mark where the function is defined (at the left endpoint of each step). At $x = 3$: right-continuous (the step continues to the right), but not left-continuous (the value jumps from $2$ to $3$).

The floor function is right-continuous at every integer but not left-continuous. So it is discontinuous at every integer. Between integers — say at x = 3.7 — both one-sided limits equal 3, which is the function's value, so it is continuous there.

One-sided continuity becomes especially important when a function is defined on a closed interval [a, b]. At the left endpoint a, you can only approach from the right (there is nothing to the left), so you only need right-continuity there. At the right endpoint b, you only need left-continuity. This is exactly what happens next.

Continuity on an interval

A function that is continuous at a single point is useful, but the real power comes when a function is continuous across an entire stretch of the number line.

Continuity on an interval

  • f is continuous on the open interval (a, b) if f is continuous at every point in (a, b).
  • f is continuous on the closed interval [a, b] if:
    1. f is continuous on (a, b),
    2. f is right-continuous at a: \displaystyle\lim_{x \to a^+} f(x) = f(a),
    3. f is left-continuous at b: \displaystyle\lim_{x \to b^-} f(x) = f(b).

The boundary treatment is natural. At the endpoints, the interval only extends in one direction, so only the corresponding one-sided limit needs to match.

A function that is continuous on (-\infty, \infty) — the entire real line — is called continuous everywhere or simply a continuous function. Polynomials like x^2 - 3x + 7 are continuous everywhere. So are \sin x, \cos x, and e^x. These are the functions whose graphs have no breaks at all.

Checking continuity: a worked example

Time to apply the definition to an actual function and verify continuity step by step.

Example 1: Is f(x) = 2x + 3 continuous at x = 1?

Step 1. Check that f(1) is defined.

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

Why: the function is a polynomial — it is defined everywhere, so this step is automatic.

Step 2. Compute \lim_{x \to 1} f(x).

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

Why: for a polynomial, you can compute the limit by direct substitution. There is no division by zero, no indeterminate form — just plug in.

Step 3. Compare the limit with the function value.

\lim_{x \to 1} f(x) = 5 = f(1)

Why: the limit equals the value, so all three conditions are satisfied.

Step 4. Conclude.

f(x) = 2x + 3 is continuous at x = 1.

Result: Continuous at x = 1, and by the same argument at every real number. A straight line has no holes, no jumps, no breaks — anywhere.

The line $y = 2x + 3$. At $x = 1$, the function value is $5$, and the graph passes smoothly through $(1, 5)$. No hole, no jump — continuous.

The graph confirms what the algebra said. The line passes through (1, 5) without any disruption. Every linear function y = mx + c is continuous everywhere, because the limit at any point is always just the value at that point.

That example was intentionally simple. The three-condition check is trivial for polynomials — but the same framework applies to the hard cases. Here is one.

Example 2: Checking a piecewise function at the boundary

Consider the piecewise function:

f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 3x - 1 & \text{if } x \geq 2 \end{cases}

Is f continuous at x = 2?

Step 1. Check that f(2) is defined.

Since 2 \geq 2, use the second piece: f(2) = 3(2) - 1 = 5.

Why: always check which piece of the definition applies at the point in question.

Step 2. Compute the left-hand limit.

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 + 1) = 4 + 1 = 5

Why: for x < 2, the function equals x^2 + 1. As x approaches 2 from the left, this approaches 5.

Step 3. Compute the right-hand limit.

\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 1) = 6 - 1 = 5

Why: for x \geq 2, the function equals 3x - 1. As x approaches 2 from the right, this approaches 5.

Step 4. Since both one-sided limits equal 5, the two-sided limit exists and equals 5. Compare with the function value:

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

Why: the left and right limits agree with each other and with the function value. All three conditions hold.

Result: f is continuous at x = 2. The two pieces join seamlessly.

The two pieces of $f$ — the parabola $x^2 + 1$ for $x < 2$ and the line $3x - 1$ for $x \geq 2$ — meet at the point $(2, 5)$. Because both pieces arrive at the same value, the join is seamless and $f$ is continuous.

The geometry confirms the algebra: the parabola and the line meet at exactly the same point, (2, 5). There is no gap, no jump, and no hole at the boundary. If the two pieces had arrived at different heights — say if the line piece had been 3x instead of 3x - 1, giving a right-hand limit of 6 instead of 5 — the function would have a jump discontinuity at x = 2.

Examples of continuous and discontinuous functions

With the definition in hand, here is a catalogue of the standard cases.

Continuous everywhere

Continuous on their domain, discontinuous outside

Discontinuous at specific points

The sign function $\text{sgn}(x)$. The function jumps from $-1$ to $+1$ at the origin, with the value at $x = 0$ defined as $0$. The left limit ($-1$), the right limit ($+1$), and the function value ($0$) are all different — three things that should agree, and none of them do.

Common confusions

Going deeper

If you came here to understand what continuity means and how to check it, you have everything you need — you can stop here. The rest is for readers who want the rigorous epsilon-delta version, the formal reason behind the three conditions, and a glimpse of what continuity makes possible.

The epsilon-delta definition

The definition above used limits, and limits themselves have a rigorous formulation in terms of \varepsilon (epsilon) and \delta (delta). When you unwind the limit in the continuity definition, you get:

f is continuous at x = a if for every \varepsilon > 0, there exists a \delta > 0 such that

|x - a| < \delta \implies |f(x) - f(a)| < \varepsilon

Read it carefully. It says: no matter how small a tolerance \varepsilon you set for the output, there is a window of width 2\delta around a in the input such that every x in that window produces an f(x) within \varepsilon of f(a).

This is the same three-condition check, compressed into one formal sentence. The existence of f(a) is implicit (you are writing f(a), so it had better exist). The existence of the limit is guaranteed by the \delta that works for every \varepsilon. And the equality of the limit and the value is built into the |f(x) - f(a)| < \varepsilon — the function values near a cluster around f(a) itself, not around some other number.

Verifying epsilon-delta for f(x) = 3x + 1 at x = 2

Take \varepsilon > 0. You need |f(x) - f(2)| < \varepsilon, i.e., |3x + 1 - 7| < \varepsilon, i.e., |3x - 6| < \varepsilon, i.e., 3|x - 2| < \varepsilon. This holds whenever |x - 2| < \varepsilon/3. So choose \delta = \varepsilon/3. For every \varepsilon, this \delta works, and the definition is satisfied. The function is continuous at x = 2.

This is the skeleton of every epsilon-delta proof of continuity. You start from the inequality |f(x) - f(a)| < \varepsilon, manipulate it until you get |x - a| < \text{something}, and then declare that "something" to be your \delta.

Why continuity matters: the big theorems

Continuity is not just a hygiene check. It is the hypothesis of some of the most important theorems in calculus.

The Intermediate Value Theorem (IVT). If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then there is some c in (a, b) where f(c) = 0. In plain language: a continuous function that starts positive and ends negative (or vice versa) must cross zero somewhere in between. It cannot "jump over" zero without passing through it — because a continuous function has no jumps.

This is how you prove that equations have solutions without actually solving them. For instance, x^3 - x - 1 = 0 has a root between 1 and 2, because f(1) = -1 < 0 and f(2) = 5 > 0, and f is a polynomial (hence continuous).

The function $f(x) = x^3 - x - 1$ is continuous. Since $f(1) = -1$ (below zero) and $f(2) = 5$ (above zero), the Intermediate Value Theorem guarantees a root between $1$ and $2$. The graph crosses zero near $x \approx 1.325$.

The Extreme Value Theorem. If f is continuous on a closed interval [a, b], then f attains both a maximum and a minimum value somewhere in that interval. A continuous function on a closed interval cannot escape to infinity or oscillate without settling — it is bounded and achieves its bounds.

Both theorems fail the moment you drop continuity. A function with a single jump discontinuity can cross from positive to negative without ever being zero (it jumps over zero). A function with an infinite discontinuity can blow up without having a maximum. Continuity is not decoration — it is the structural condition that makes calculus work.

Where this leads next

You now know what continuity means at a point, on an interval, and from one side. The next articles build on this.