Continuity

Continuity is the mathematical way of saying a function has "no breaks, no holes, and no jumps" — you could draw its graph without lifting your pen. That intuition is genuinely useful, but it hides a subtle truth: what looks smooth to the eye can hide a puncture, and what the eye cannot see, calculus can. Continuity is the property that lets limits, derivatives, and integrals behave the way we expect, and it is the quiet assumption behind nearly every theorem in a first calculus course.

In this page you will learn the precise, three-part definition of continuity, meet the three main types of discontinuity, and see why the Intermediate Value Theorem — a formal statement of "to get from here to there, you must pass through everything in between" — is such a powerful tool.

Learning Objectives

• State the three conditions a function must satisfy to be continuous at a point.
• Test a function for continuity at a point using limits.
• Classify discontinuities as removable, jump, or infinite.
• State and apply the Intermediate Value Theorem (IVT).
• Understand the historical need that pushed Bolzano and Cauchy to define continuity rigorously.

Quick Answer

A function $f$ is continuous at a point $x = a$ if three things all hold: (1) $f(a)$ is defined, (2) the limit $\lim_{x \to a} f(x)$ exists, and (3) that limit equals $f(a)$. In one line: $\lim_{x \to a} f(x) = f(a)$. If any condition fails, $f$ has a discontinuity at $a$. Discontinuities come in three flavors: removable (a hole you could "plug"), jump (left and right limits disagree), and infinite (the function blows up, as at a vertical asymptote). A function continuous on a closed interval satisfies the Intermediate Value Theorem: it takes every value between $f(a)$ and $f(b)$ somewhere in between.

Where It Came From

For over a century after Newton and Leibniz invented calculus in the 1680s, mathematicians used "continuous" loosely — a curve was continuous if it was given by a single formula, or if it had no visible gaps. This vagueness caused real trouble. Results that seemed obvious kept producing paradoxes, and no one could say exactly why a formula-based curve should behave nicely. The concept was doing heavy lifting in proofs without ever being pinned down.

The Bohemian priest and mathematician Bernard Bolzano, around 1817, was the first to give a definition close to the modern one. In a paper with the imposing title Rein analytischer Beweis ("Purely Analytic Proof"), he wanted to prove the Intermediate Value Theorem without appealing to geometry or motion — purely from the definition of the numbers involved. To do that he needed to say precisely what "continuous" meant, and he defined it in terms of the function's values getting arbitrarily close: $f(x)$ differs from $f(a)$ by as little as we please when $x$ is close enough to $a$.

Bolzano's work was published in an obscure venue and went largely unnoticed. A few years later, the French analyst Augustin-Louis Cauchy, in his hugely influential 1821 textbook Cours d'analyse, independently gave essentially the same definition and put it at the center of how calculus was taught. Cauchy's authority made the idea stick. Later, Karl Weierstrass sharpened it into the fully symbolic epsilon–delta ($\varepsilon$–$\delta$) definition we use today. The motivation throughout was the same: to replace "you can draw it without lifting your pen" — a claim about ink and hands — with a statement about numbers that a proof could actually rest on.

The Three Conditions for Continuity

To be continuous at a point $x = a$, a function must pass all three of these tests:

1. $f(a)$ exists — the point is actually in the domain; there is a value there.
2. $\lim_{x \to a} f(x)$ exists — the function approaches a single finite value from both sides (so the left-hand and right-hand limits agree).
3. $\lim_{x \to a} f(x) = f(a)$ — the value the function approaches matches the value it takes.

All three are needed. A function can be defined at $a$ yet have no limit (condition 2 fails), or have a perfectly good limit that simply doesn't match the value plotted there (condition 3 fails).

Worked example. Test whether

$f(x) = \frac{x^2 - 9}{x - 3}$

is continuous at $x = 3$.

• Condition 1: $f(3)$ requires dividing by $3 - 3 = 0$, so $f(3)$ is undefined. Condition 1 already fails.

The limit, however, exists. Factor: $\dfrac{x^2 - 9}{x - 3} = \dfrac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$, so

$\lim_{x \to 3} f(x) = 3 + 3 = 6.$

The limit is a clean 6, but the function has no value at $x = 3$. This is a hole — a removable discontinuity, which we meet next.

Types of Discontinuity

When continuity fails, the way it fails tells us a lot. There are three standard types.

Removable Discontinuity (a hole)

The two-sided limit exists, but either $f(a)$ is undefined or it disagrees with the limit. It is called "removable" because we could redefine (or define) $f(a)$ to equal the limit and repair the break with a single point.

Worked example. The function above, $f(x) = \frac{x^2-9}{x-3}$, has a removable discontinuity at $x = 3$. Defining a new function

$g(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 6, & x = 3 \end{cases}$

makes $g$ continuous everywhere: we simply plugged the hole with the value the limit was already pointing at.

Jump Discontinuity

The left-hand and right-hand limits both exist but are different numbers, so the two-sided limit does not exist. The graph "jumps" from one height to another. These are common in step functions and piecewise definitions.

Worked example. Consider

$f(x) = \begin{cases} x + 1, & x < 2 \\ x + 4, & x \geq 2 \end{cases}$

From the left: $\lim_{x \to 2^-} f(x) = 2 + 1 = 3$. From the right: $\lim_{x \to 2^+} f(x) = 2 + 4 = 6$. Since $3 \neq 6$, the limit does not exist and the function jumps by 3 units at $x = 2$. No single redefinition can fix this — the two sides genuinely disagree.

Infinite Discontinuity

At least one one-sided limit is infinite. The function shoots off toward $+\infty$ or $-\infty$, typically at a vertical asymptote.

Worked example. For $f(x) = \dfrac{1}{x - 2}$, as $x \to 2^+$ the denominator is a tiny positive number, so $f(x) \to +\infty$; as $x \to 2^-$ it is a tiny negative number, so $f(x) \to -\infty$. Because the values grow without bound, there is no way to assign a finite value at $x = 2$ that makes $f$ continuous. This is an infinite (essential) discontinuity.

The Intermediate Value Theorem

The Intermediate Value Theorem (IVT) captures the heart of what continuity buys you.

If $f$ is continuous on a closed interval $[a, b]$, and $N$ is any number strictly between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a, b)$ with $f(c) = N$.

Informally: a continuous function cannot skip values. To move from height $f(a)$ to height $f(b)$ without lifting the pen, it must pass through every height in between. Note the theorem requires continuity — this is exactly why the rigorous definition mattered to Bolzano.

Worked example — root finding. Show that $f(x) = x^3 - x - 1$ has a root between $x = 1$ and $x = 2$.

• $f$ is a polynomial, so it is continuous everywhere, including on $[1, 2]$.
• $f(1) = 1 - 1 - 1 = -1$ (negative).
• $f(2) = 8 - 2 - 1 = 5$ (positive).

Since $0$ lies between $-1$ and $5$, the IVT guarantees some $c$ in $(1, 2)$ with $f(c) = 0$. The theorem promises the root exists; narrowing it down (say to $c \approx 1.325$) is the job of numerical methods like bisection, which are built directly on this idea.

Real-World Applications

• Root-finding algorithms: Bisection, used everywhere from calculators to engineering solvers, relies on the IVT — bracket a sign change, then repeatedly halve the interval.
• Temperature and physics: If a rod is $20°$C at one end and $80°$C at the other, and temperature varies continuously, every value between $20°$ and $80°$ occurs somewhere along it. This is a direct IVT statement.
• Economics: A continuous cost or demand curve lets analysts guarantee a "break-even" price exists between a loss and a profit region.
• Computer graphics and animation: Continuous (and smoothly continuous) functions ensure objects move without teleporting or flickering between frames.
• Control systems: Engineers assume sensor-to-output mappings are continuous so that small input changes don't cause sudden dangerous jumps.

Common Mistakes

1. Thinking "the limit exists" is enough for continuity. Why it's wrong: A removable discontinuity has a perfectly good limit but no matching value. All three conditions must hold. Correction: Always check that $f(a)$ exists and equals the limit — not just that the limit exists.

2. Assuming a function defined by a formula is automatically continuous everywhere. Why it's wrong: Formulas like $\frac{1}{x}$ or $\tan x$ have points where the denominator is zero and continuity fails. Correction: Check the domain first; a function can only be continuous at points where it is actually defined.

3. Applying the IVT without checking continuity, or expecting it to find every root. Why it's wrong: The IVT needs continuity on a closed interval, and it only guarantees at least one value is hit — it does not count roots or locate them exactly. Correction: Confirm continuity on $[a, b]$, and remember the IVT is an existence statement, not a counting or locating tool.

Comparison and Connections

Continuity, limits, and differentiability are closely related but distinct. A limit can exist where the function isn't even defined; continuity is stronger; differentiability is stronger still.

Property	What it requires	Example that has it but not the next
Limit exists at $a$	Left = right limit (finite)	$\frac{x^2-9}{x-3}$ at $x=3$: limit is 6, but not continuous (hole)
Continuous at $a$	Limit exists and equals $f(a)$	$\lvert x \rvert$ at $x=0$: continuous but not differentiable (sharp corner)
Differentiable at $a$	Continuous and has a well-defined tangent slope	$x^2$ everywhere

Key rule of thumb: differentiable $\Rightarrow$ continuous, but continuous $\not\Rightarrow$ differentiable. The absolute value function is the classic counterexample: continuous everywhere, but with a corner at $0$ where no single slope exists.

Practice Questions

Recall

State the three conditions for $f$ to be continuous at $x = a$. Answer: (1) $f(a)$ is defined; (2) $\lim_{x \to a} f(x)$ exists; (3) $\lim_{x \to a} f(x) = f(a)$.

Understanding

Classify the discontinuity of $f(x) = \dfrac{x-4}{x^2 - 16}$ at $x = 4$ and at $x = -4$. Guidance: Factor the denominator: $x^2 - 16 = (x-4)(x+4)$, so $f(x) = \frac{1}{x+4}$ for $x \neq 4$. At $x = 4$ the factor cancels, giving a finite limit of $\frac{1}{8}$ — a removable discontinuity. At $x = -4$ the denominator still vanishes and the function blows up — an infinite discontinuity.

Application

Use the IVT to show that $\cos x = x$ has a solution in $[0, 1]$. Guidance: Let $g(x) = \cos x - x$, continuous everywhere. Then $g(0) = 1 - 0 = 1 > 0$ and $g(1) = \cos 1 - 1 \approx 0.540 - 1 = -0.460 < 0$. Since $g$ changes sign, the IVT guarantees a $c$ in $(0,1)$ with $g(c) = 0$, i.e. $\cos c = c$.

Analysis

Is it possible for a function to be continuous at exactly one point? Explain. Guidance: Yes. Consider $f(x) = x^2$ if $x$ is rational and $f(x) = 0$ if $x$ is irrational. Away from $0$, values near any point include both branches, which disagree, so continuity fails. At $x = 0$ both branches give values squeezing to $0 = f(0)$, so $f$ is continuous only there. This shows continuity is genuinely a point-by-point property, not something that must hold on whole intervals.

FAQ

Is a function continuous at an endpoint of its domain? Yes, using one-sided limits. At a left endpoint we only require $\lim_{x \to a^+} f(x) = f(a)$; the "other side" isn't part of the domain, so we don't demand it.

Does "the limit exists" mean the function is continuous? No. The limit can exist while the function is undefined there or takes a different value — that's a removable discontinuity. Continuity needs the limit and a matching function value.

What's the difference between removable and jump discontinuities? For a removable discontinuity the two-sided limit exists (both sides agree) but the function value is missing or wrong. For a jump, the left and right limits are different numbers, so no two-sided limit exists and nothing can "plug" it.

Can the Intermediate Value Theorem tell me how many roots a function has? No. It only guarantees at least one value is attained. A function might cross a level once, three times, or infinitely often — the IVT doesn't distinguish. It is purely an existence guarantee.

If a function is continuous, is it automatically differentiable? No. Continuity is necessary for differentiability but not sufficient. $f(x) = \lvert x \rvert$ is continuous at $0$ but has a corner there, so it has no derivative at that point.

Quick Revision

• Definition: $f$ is continuous at $a$ iff $\lim_{x \to a} f(x) = f(a)$ (value defined, limit exists, they match).
• Removable: limit exists, value missing/wrong — a hole; can be patched.
• Jump: left limit $\neq$ right limit — two-sided limit fails.
• Infinite: function $\to \pm\infty$; vertical asymptote.
• IVT: $f$ continuous on $[a,b]$ hits every value $N$ between $f(a)$ and $f(b)$. Sign change $\Rightarrow$ a root exists.
• Hierarchy: differentiable $\Rightarrow$ continuous $\Rightarrow$ limit exists (arrows do not reverse).
• History: Bolzano (1817) and Cauchy (1821) replaced "no breaks" with a limit-based definition; Weierstrass gave the $\varepsilon$–$\delta$ form.

Related Topics

Prerequisites

• Limits — continuity is defined entirely in terms of limits.
• Calculus overview — where continuity fits in the bigger picture.

Related Topics

• Derivatives — differentiability builds on continuity.

Next Topics

• The Extreme Value Theorem and applications of continuity to optimization.
• Numerical root-finding (bisection) as a direct application of the IVT.