Continuity at a Point

Consider a function that computes a package's shipping cost from its weight. Suppose the cost jumps from $5.00 to $5.01 for an infinitesimal increase in weight. This happens at exactly 1 kg. Something feels wrong, even though the function is perfectly well-defined at that point. Continuity at a point is the precise idea that makes this discomfort mathematically rigorous. It says a function's value at a point must agree with the trend of its values nearby. No jump, hole, or mismatch is allowed. This concept matters for an important reason. Almost every theorem in analysis that lets us compute presupposes continuity as a baseline regularity condition. This includes the Intermediate Value Theorem, the Extreme Value Theorem, differentiability, and integrability. Continuity is the bridge between two ideas. One is the purely local, dynamic notion of a limit (what happens as you approach a point). The other is the static, algebraic notion of a function's value (what happens at the point). Confusing these two ideas is one of the most common errors a beginning analysis student makes. The big picture is this: a limit can exist at a point even if the function is undefined there. It can also exist if the function is defined but with a different, "wrong" value. Continuity closes that gap. It demands three things simultaneously — that the function be defined at the point, that its limit exist there, and that the two coincide. This article isolates continuity as a pointwise property. It distinguishes continuity sharply from mere limit existence. It also gives both the limit-based and the direct epsilon-delta formulations, proving they are equivalent. What follows is a full treatment. It includes the formal definition stated two equivalent ways, and a rigorous proof of their equivalence. It offers worked examples that expose the subtlety. These include a removable discontinuity where the limit exists but continuity fails. It also gives a counterexample showing why every clause of the definition is load-bearing. Finally, it maps out where this pointwise notion grows into continuity on whole sets. That is the subject of Continuity on Intervals and Sets and The Algebra of Continuous Functions.

Key takeaways

Continuity at $a$ requires three things at once: $f(a)$ is defined, $\lim_{x\to a} f(x)$ exists, and the two are equal.
A limit can exist at a point where $f$ is undefined, or where $f$ is defined but takes a different value — neither case is continuity.
The epsilon-delta form: for every $\varepsilon>0$ there is $\delta>0$ so that $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\varepsilon$ — note $x=a$ is now allowed.
Continuity is a genuinely local/pointwise property: it can hold at one point of a function's domain and fail at another.
Removable discontinuities are exactly the case "limit exists, but doesn't match (or isn't) the function's value" — the mismatch is fixable by redefinition.
Key idea: continuity glues the dynamic behavior of $f$ near $a$ to the static value $f(a)$ , with no gap permitted.

Where this lives today

Continuity at a point is the atomic unit from which every stronger notion of regularity in analysis is built. In real analysis, it underlies the Sequential Characterization of Continuity (an equivalent formulation via convergent sequences, useful for proving discontinuity by exhibiting a "bad" sequence) and the extension to whole-domain continuity in Continuity on Intervals and Sets, which powers the Intermediate Value Theorem and Extreme Value Theorem. In physics and engineering, continuity at a point models the physically reasonable requirement that quantities like temperature, displacement, or voltage cannot jump instantaneously without infinite energy or force — control theory and signal processing explicitly track discontinuities (jumps, spikes) as flags for unmodeled physical behavior. In numerical analysis, root-finding algorithms like bisection depend on continuity to guarantee that a sign change implies a root nearby. In probability and statistics, cumulative distribution functions are always right-continuous by convention, and continuity at a point of a CDF characterizes whether the underlying random variable has an atom (positive probability mass) there. In computer science and cryptography, notions of "continuity" generalize to Lipschitz and uniform continuity conditions used in numerical stability analysis and in complexity-theoretic treatments of computable analysis, where "continuous at a point" becomes a computability requirement on real-number algorithms.

Ask the tutor: If $f$ and $g$ are both continuous at $a$ , why must $f+g$ and $f\cdot g$ also be continuous at $a$ ?

f: D \to \mathbb{R}

Intuition

Picture the graph of $f$ near the point $(a, f(a))$ . The limit statement $\lim_{x\to a} f(x) = L$ only cares about a punctured neighborhood of $a$ : what the curve does as $x$ sneaks up on $a$ from either side, never mind what happens (or doesn't happen) exactly at $x=a$ . Continuity adds the missing ingredient — it insists that the destination the curve is heading toward, $L$ , is also where the curve actually lands, namely $f(a)$ . This is why a graph "with no hole and no jump" is the right cartoon: a hole means the limit exists but $f(a)$ is either undefined or defined to a different value; a jump means the limit fails to exist at all (the left and right approach different values), so there is nothing for $f(a)$ to match. Continuity rules out both defects at the single point $a$ . The epsilon-delta form sharpens this picture into a control statement: no matter how tight a tolerance $\varepsilon$ an adversary demands around the target value $f(a)$ , you can find a window of $x$ -values around $a$ (radius $\delta$ ) all of which land inside that tolerance — and this window is allowed to include $a$ itself, where the claim is automatic.

Figure

Three behaviors at

x=a

: continuous, removable gap, and jump

f: D \to \mathbb{R}

x=a

a \in D

\delta = \delta_0

This corollary is a useful sanity check: continuity at isolated points carries no real information about $f$ 's "smoothness," which is exactly why deeper theorems (Intermediate Value, differentiability, etc.) are usually stated for continuity on intervals rather than at isolated points — a distinction developed in Continuity on Intervals and Sets.

Worked examples

Example 1 (a removable discontinuity, and how to fix it). Define $f: \mathbb{R} \to \mathbb{R}$ by

f(x) = \begin{cases} \dfrac{x^3 - 8}{x-2} & x \ne 2 \\[4pt] 5 & x = 2. \end{cases}

We check continuity at $a = 2$ using the limit form. First, factor: for $x \ne 2$ ,

\frac{x^3-8}{x-2} = \frac{(x-2)(x^2+2x+4)}{x-2} = x^2 + 2x + 4,

valid because $x \ne 2$ lets us cancel the nonzero factor $(x-2)$ . By the Algebra of Limits (limits of polynomials, built from the limit of the identity function and constant functions via sums and products),

\lim_{x\to 2} \left(x^2+2x+4\right) = 2^2 + 2\cdot 2 + 4 = 4+4+4=12.

Since $f(x) = x^2+2x+4$ for all $x\ne2$ , and the limit as $x\to 2$ only depends on values with $x \ne 2$ (recall the punctured condition $0<|x-2|<\delta$ in the definition of a limit), we get

\lim_{x\to 2} f(x) = 12.

Now compare to $f(2)$ : by definition, $f(2) = 5$ . Since

\lim_{x\to 2} f(x) = 12 \ne 5 = f(2),

condition (A) fails, so $f$ is discontinuous at $2$ . This is a removable discontinuity: the limit exists (it equals $12$ ), but the function's assigned value ( $5$ ) doesn't match it. If we define $g(x) = f(x)$ for $x \ne 2$ and $g(2) := 12$ , then $\lim_{x\to 2} g(x) = 12 = g(2)$ , so $g$ is continuous at $2$ by the theorem. This distinguishes removable discontinuities sharply from jump or essential discontinuities, where no single redefinition at $a$ can restore continuity because the limit itself fails to exist.

Example 2 (continuity verified directly by epsilon-delta, no limit laws). Let $f(x) = 3x^2 - x$ on $\mathbb{R}$ , and verify continuity at $a = -1$ directly from condition (B), without invoking the algebra of limits. We have $f(-1) = 3(1) - (-1) = 3+1 = 4$ . Fix $\varepsilon > 0$ . We must find $\delta > 0$ such that $|x+1| < \delta \implies |f(x) - 4| < \varepsilon$ . Compute:

f(x) - 4 = 3x^2 - x - 4 = (x+1)(3x-4).

(Check: $(x+1)(3x-4) = 3x^2 -4x+3x-4 = 3x^2 - x - 4$ . ✓.) So $|f(x)-4| = |x+1|\,|3x-4|$ . We need to bound $|3x-4|$ in terms of a fixed neighborhood of $-1$ . Restrict attention to $|x+1| < 1$ , i.e., $-2 < x < 0$ . On this range,

|3x - 4| \le 3|x| + 4 < 3\cdot 2 + 4 = 10,

using $|x| < 2$ for $x \in (-2,0)$ (since $|x+1|<1 \Rightarrow |x|\le |x+1|+1 < 2$ by the triangle inequality). Thus, whenever $|x+1| < 1$ ,

|f(x) - 4| = |x+1|\,|3x-4| < 10\,|x+1|.

Given $\varepsilon > 0$ , choose

\delta = \min\left(1, \frac{\varepsilon}{10}\right).

Then for $|x+1| < \delta$ : first $|x+1|<1$ holds, so the bound $|3x-4|<10$ applies, and also $|x+1| < \varepsilon/10$ , giving

|f(x)-4| < 10 \cdot \frac{\varepsilon}{10} = \varepsilon.

This verifies condition (B) at $a=-1$ for every $\varepsilon>0$ , so $f$ is continuous at $a=-1$ . (The same computation, done symbolically at a general point $a$ , is exactly the content of "polynomials are continuous everywhere," a fact used freely once The Algebra of Continuous Functions is established — but here we did it from scratch to show the direct $\varepsilon$ - $\delta$ mechanics.)

a \in D

Scope, edge cases & common traps

Situation	What happens	Is $f$ continuous at $a$ ?
$\lim_{x\to a} f(x)$ exists, $f(a)$ defined, values equal	limit form (A) holds	Yes
$\lim_{x\to a} f(x)$ exists, $f(a)$ defined, values differ	removable discontinuity	No
$\lim_{x\to a} f(x)$ exists, $f(a)$ undefined ( $a \notin D$ )	not in scope of the definition	Undefined question — need $a \in D$
$\lim_{x\to a} f(x)$ does not exist (e.g. left/right limits differ)	jump discontinuity	No, regardless of $f(a)$
$a$ is an isolated point of $D$	condition (B) holds vacuously	Yes, automatically
$f$ oscillates without settling (e.g. $\sin(1/x)$ near $0$ , if $0\in D$ )	limit fails to exist at all	No

Common trapThe classic error: treating "limit exists" as if it already meant "continuous"

A very common mistake is to compute $\lim_{x\to a} f(x)$ , find that it exists and equals some number $L$ , and conclude "therefore $f$ is continuous at $a$ " — without ever checking that $f(a)$ is even defined, let alone that $f(a) = L$ . Example 1 above is a direct demonstration: the limit at $2$ exists and equals $12$ , yet $f$ is discontinuous at $2$ because $f(2) = 5 \ne 12$ . The correct logical order is always: (i) confirm $a \in D$ so $f(a)$ exists; (ii) confirm $\lim_{x\to a} f(x)$ exists; (iii) confirm the two values match. Skipping step (iii) — or worse, skipping (i) — silently converts a true statement about limits into a false claim about continuity.

The mathematician behind it

Origins · 1 minKarl Weierstrass and how this was discovered

Karl Weierstrass

1815–1897 · Prussia/German Empire, foundations of analysis

Karl Weierstrass is the figure most responsible for the modern $\varepsilon$ – $\delta$ formulation used throughout this article. Trained originally in law before turning to mathematics, he spent years as a secondary-school teacher in relative obscurity, teaching gymnastics and penmanship alongside mathematics, before a single paper on abelian functions (1854) brought him sudden recognition and a position at the University of Berlin. There he built, through lecture courses rather than published treatises, the rigorous arithmetization of analysis: precise definitions of limit, continuity, and convergence purged of appeals to infinitesimals or geometric intuition. His students, including Sofia Kovalevskaya (one of the first women to earn a doctorate in mathematics, under his direct supervision) and Georg Cantor, carried this rigor forward. Weierstrass is also famous for constructing a function continuous everywhere but differentiable nowhere, a startling counterexample that only became fully meaningful once continuity itself had been pinned down with the precision this article states.

History

- 1817 — Bernard Bolzano gives an early, largely unnoticed

\varepsilon

-style definition of continuity in a proof of the Intermediate Value Theorem.

- 1821 — Augustin-Louis Cauchy publishes Cours d'Analyse, defining continuity via infinitesimal increments, a major step toward rigor but still tied to intuitive infinitesimals.

- 1850s–1860s — Karl Weierstrass, in Berlin lectures, formulates the modern

\varepsilon

–

\delta

definition of both limit and continuity at a point, purging infinitesimal language.

- 1872 — Weierstrass presents his continuous-but-nowhere-differentiable function, showing continuity alone guarantees far less regularity than intuition suggests.

- 1900s — The

\varepsilon

–

\delta

formulation becomes the standard in analysis textbooks worldwide, cementing the pointwise definition used today.

Why does continuity require

a

to be in the domain, but the limit definition does not?

The limit $\lim_{x\to a} f(x)$ is fundamentally a statement about the trend of $f$ 's values as $x$ approaches $a$ , deliberately excluding $x=a$ via the strict inequality $0<|x-a|<\delta$ . This lets limits be discussed even when $f(a)$ is undefined, as with $\frac{\sin x}{x}$ at $x=0$ . Continuity, by contrast, is a comparison: it asks whether the trend (the limit) matches the actual assigned value $f(a)$ . If $f(a)$ doesn't exist, there is nothing to compare the limit to, so the question "is $f$ continuous at $a$ " is meaningless rather than false. Requiring $a \in D$ makes the definition well-posed rather than vacuous or undefined.

Can a function be continuous at exactly one point and nowhere else?

Yes. A classic example is $f(x) = x$ if $x$ is rational and $f(x)=0$ if $x$ is irrational, defined on all of $\mathbb{R}$ . One can show, using the density of both rationals and irrationals in every interval, that $f$ is continuous only at $a=0$ and discontinuous at every other point. This demonstrates that continuity truly is a pointwise property — it need not hold uniformly across a domain, and asking "is $f$ continuous" without specifying a point (or a whole set) is often an incomplete question until the stronger notion in Continuity on Intervals and Sets is invoked.

Is it possible for

\lim_{x\to a} f(x)

to exist while

f

is discontinuous at

a

Yes, and this is precisely the removable discontinuity case worked out in Example 1. The limit existing only guarantees that $f$ 's values near $a$ settle down to some number $L$ ; it says nothing about $f(a)$ itself, which is an independent piece of data. If $f(a)$ happens to be undefined (so $a\notin D$ ) or defined but unequal to $L$ , continuity fails even though the limit is perfectly well-behaved. This is the single most important distinction this article draws.

Does continuity at a point say anything about differentiability there?

No — continuity is strictly weaker. Every differentiable function is continuous at each point where it is differentiable (a fact usually proved right after continuity is developed, using the definition of the derivative as a limit of difference quotients). But the converse fails dramatically: $f(x)=|x|$ is continuous at $0$ but not differentiable there, and Weierstrass's 1872 function is continuous at every point of $\mathbb{R}$ yet differentiable at none. Continuity only forbids jumps and gaps in value; it says nothing about the existence of a well-defined tangent slope.

How do you prove a function is discontinuous at a point, using epsilon-delta directly?

To disprove continuity via (B) directly, you must exhibit a single "bad" $\varepsilon_0>0$ for which no $\delta>0$ works: for every candidate $\delta$ , you exhibit some $x$ with $|x-a|<\delta$ yet $|f(x)-f(a)|\ge \varepsilon_0$ . This is logically the negation of the quantifier string $\forall\varepsilon\,\exists\delta\,\forall x(\dots)$ , which becomes $\exists \varepsilon\,\forall\delta\,\exists x(\dots \text{ and } |f(x)-f(a)|\ge\varepsilon)$ . In practice this is often easier via the (equivalent, later-developed) Sequential Characterization of Continuity: find one sequence $x_n \to a$ with $f(x_n) \not\to f(a)$ .

Is continuity at a point the same as one-sided continuity from the left and right both holding?

Yes, provided $a$ is a two-sided limit point of the domain (i.e., both sides are represented near $a$ ). In that setting, $\lim_{x\to a} f(x)$ exists and equals $f(a)$ if and only if both one-sided limits exist, are equal to each other, and equal $f(a)$ . If $a$ is an endpoint of the domain (only one side is present, e.g. $a$ is the left endpoint of $[a,b]$ ), continuity at $a$ reduces to the single relevant one-sided limit condition, since the epsilon-delta condition is only ever tested against $x\in D$ .

Why do we quantify "for every

\varepsilon

" before "there exists

\delta

", and why does that order matter?

The order encodes who moves first: an adversary picks an arbitrarily tight tolerance $\varepsilon>0$ around the target value $f(a)$ , and only afterward do you get to respond with a window size $\delta$ (which is allowed, and generally needs, to depend on $\varepsilon$ ). Reversing the order — "there exists $\delta$ such that for every $\varepsilon$ ..." — would demand a single $\delta$ that works uniformly for all tolerances simultaneously, an entirely different and generally false statement (it would force $f$ to be locally constant). Respecting this quantifier order is essential; swapping it is one of the most common logical errors in first exposure to $\varepsilon$ – $\delta$ arguments.