The Epsilon Delta Definition of a Limit, Explained

Q: What is the epsilon delta definition of a limit?

The limit of f(x) as x approaches a equals L if, for every epsilon greater than 0, there exists a delta greater than 0 such that whenever 0 is less than the absolute value of x minus a, which is less than delta, the absolute value of f(x) minus L is less than epsilon. In words, any tolerance you demand on the output can be met by restricting the input to a small enough punctured interval around a.

Q: Why do we need the formal definition of a limit?

Because the intuitive version, saying f(x) gets close to L as x gets close to a, uses undefined words and hides a contradiction. Early calculus divided by an infinitesimal quantity and then discarded it as zero in the same computation. Berkeley attacked exactly this in 1734. The epsilon delta definition replaces motion and infinitesimals with inequalities between ordinary real numbers, so no quantity has to be nonzero and zero at once.

Q: What does it mean for a limit to exist?

It means there is a single number L that works for every epsilon. If two different sequences approaching a drive f towards two different values, no single L can satisfy the definition for small epsilon, and the limit does not exist. That is why the limit of sin of 1 over x as x tends to 0 fails, and why 1 over x has no limit at 0 even though each one-sided behaviour is describable.

Q: How do you do an epsilon delta proof?

Work backwards first. Start from the target inequality, the absolute value of f(x) minus L is less than epsilon, and manipulate it until it is expressed in terms of the absolute value of x minus a. That scratch work reveals the required delta. Then write the proof forwards: let epsilon be given, define delta explicitly, assume the input condition, and derive the output condition. For nonlinear functions you usually restrict delta to at most 1 first in order to bound the extra factor.

Q: What is an infinitesimal?

A quantity greater than 0 but smaller than every positive real number. No such real number exists, which is why the epsilon delta definition avoids the idea entirely. Abraham Robinson made infinitesimals rigorous in 1966 by constructing the hyperreal numbers, an ordered field extending the reals that genuinely contains them, so the older intuition was eventually vindicated rather than refuted.

In 1734 the philosopher and bishop George Berkeley published a pamphlet with a title designed to provoke: The Analyst; or, a Discourse Addressed to an Infidel Mathematician. His target was the calculus of Newton and Leibniz, which by then had been producing correct answers for roughly half a century while resting on a step that Berkeley found indefensible. In section 35 he delivered the line that has followed the subject ever since. Of the vanishing increments used in differentiation he asked: they are neither finite quantities, nor quantities infinitely small, nor yet nothing — may we not call them the ghosts of departed quantities?

He was right, and it took about 140 years to answer him. The answer is the epsilon delta definition of a limit.

Here it is, stated plainly. Let $f$ be defined on some punctured interval around $a$ . Then

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

means: for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

0 < |x - a| < \delta \quad \Longrightarrow \quad |f(x) - L| < \varepsilon.

That is the entire definition. Nothing in it moves, nothing approaches anything, and no quantity is infinitely small. There are only real numbers and inequalities. The rest of this article explains why that sentence says what you want it to say, why mathematics could not do without it, and how to use it.

The definition as a game

The cleanest way to read the quantifiers is as a two-player game between a sceptic and you.

The sceptic names a tolerance $\varepsilon > 0$ — an accuracy target for the output. Your job is to respond with a radius $\delta > 0$ and make a promise: every input within distance $\delta$ of $a$ , except possibly $a$ itself, produces an output within $\varepsilon$ of $L$ . If you can always meet the challenge, no matter how small the sceptic makes $\varepsilon$ , then $\lim_{x\to a} f(x) = L$ . If there is even one tolerance you cannot meet, the claim fails.

Three details of the game are worth flagging immediately, because each is a place students lose points.

You move second. The sceptic names $\varepsilon$ first, and $\delta$ is allowed to depend on it. Typically $\delta$ shrinks as $\varepsilon$ shrinks.
The condition is $0 < |x-a|$ , not $0 \le |x-a|$ . The point $x = a$ is deliberately excluded. This is what allows $\lim_{x\to 2}\frac{x^2-4}{x-2} = 4$ even though the function is undefined at $2$ . The limit is a statement about the neighbourhood of $a$ , never about $a$ itself.
Any smaller $\delta$ also works. If $\delta$ succeeds, so does $\delta/2$ . So you never need the largest possible $\delta$ , only some $\delta$ — which is why the proofs below are allowed to be crude.

Why the informal version is not enough

The usual first description is that $f(x)$ gets arbitrarily close to $L$ as $x$ gets close to $a$ . As a picture this is fine. As a definition it is circular: close is undefined, and gets smuggles in time and motion, which have no place in a statement about a static function.

The historical damage was worse than imprecision. Here is Newton’s calculation of the derivative of $f(x) = x^2$ , in the style of the fluxional calculus, using an increment $o$ :

\frac{(x+o)^2 - x^2}{o} = \frac{2xo + o^2}{o} = 2x + o.

The first step divides by $o$ , which requires $o \ne 0$ . The last step discards $o$ , which requires $o = 0$ . The same symbol is treated as nonzero and zero within three lines. Leibniz’s $dx$ had the identical problem in different notation. The results were right — the derivative of $x^2$ really is $2x$ — but the reasoning that produced them was not reasoning.

Berkeley’s pamphlet did not dispute the answers. It disputed the entitlement. His broader jab, that mathematicians who mocked theology for demanding faith were themselves running on faith, stung precisely because the algebra could not be defended.

From Berkeley to Weierstrass

The repair took the whole of the following century, and it did not arrive in one piece.

Bernard Bolzano gave a recognisably modern definition of continuity in a paper of 1817, but his work was published obscurely and had almost no influence at the time. Augustin-Louis Cauchy’s Cours d’Analyse of 1821 put limits rather than infinitesimals at the centre of the subject and used $\varepsilon$ and $\delta$ arguments in individual proofs — though it is worth being accurate here, since he is often credited with more than he wrote: Cauchy’s stated definition still spoke of values approaching a limit indefinitely, and he retained infinitesimals as a working language. The fully arithmetic version, with no motion and no infinitely small quantities anywhere, belongs to Karl Weierstrass and his Berlin lectures, the best-documented being the differential calculus course of 1861, whose notes were taken by Hermann Amandus Schwarz. Weierstrass’s students carried the formulation outward through the 1860s and 1870s until it became the standard.

What Weierstrass achieved was not a new theorem but a change of vocabulary that made the old theorems provable. Everything is now a statement about inequalities between finite real numbers, quantified in a fixed order. The ghosts had nowhere left to hide, because no quantity in the definition is ever asked to be both zero and nonzero.

That same rigour also made pathological examples possible. Weierstrass’s own continuous nowhere-differentiable function, which would have been incomprehensible under the old intuitions, is perfectly well-behaved as an epsilon delta object.

A complete worked proof: a linear function

Claim: $\displaystyle\lim_{x \to 3} (2x+1) = 7$ .

Scratch work. This is not the proof; it is how you find $\delta$ . Start from the conclusion you want and work backwards towards a condition on $|x-3|$ :

|f(x) - L| = |(2x+1) - 7| = |2x - 6| = 2|x-3|.

We want this to be less than $\varepsilon$ :

2|x-3| < \varepsilon \quad \Longleftrightarrow \quad |x-3| < \frac{\varepsilon}{2}.

So $\delta = \varepsilon/2$ will do. Every step here is reversible, which is what makes the backwards search legitimate.

Proof. Let $\varepsilon > 0$ be given. Choose $\delta = \varepsilon/2 > 0$ . Suppose $0 < |x-3| < \delta$ . Then

|(2x+1) - 7| = 2|x-3| < 2\delta = 2 \cdot \frac{\varepsilon}{2} = \varepsilon.

Since $\varepsilon > 0$ was arbitrary, $\lim_{x\to 3}(2x+1) = 7$ . $\blacksquare$

Note the structure, which every such proof shares: let $\varepsilon$ be given; here is $\delta$ ; assume the hypothesis; derive the conclusion. The scratch work never appears in the final write-up.

A table makes the game concrete. The last column checks the worst case at the edge of the interval.

$\varepsilon$	$\delta = \varepsilon/2$	allowed $x$	range of $2x+1$	error at $x = 3+\delta$
$1$	$0.5$	$(2.5,\,3.5)$	$(6,\,8)$	$1$
$0.1$	$0.05$	$(2.95,\,3.05)$	$(6.9,\,7.1)$	$0.1$
$0.01$	$0.005$	$(2.995,\,3.005)$	$(6.99,\,7.01)$	$0.01$
$10^{-9}$	$5 \times 10^{-10}$	—	—	$10^{-9}$

The error never reaches $\varepsilon$ on the open interval, which is exactly what the strict inequality demands.

A harder one: a nonlinear function

Claim: $\displaystyle\lim_{x \to 2} x^2 = 4$ .

Scratch work. Factor the target:

|x^2 - 4| = |x-2|\,|x+2|.

We can control $|x-2|$ directly, since that is what $\delta$ bounds. The factor $|x+2|$ is the problem: it is not constant, and near $x = 2$ it is about $4$ . The standard move is to bound it by restricting $\delta$ in advance.

Insist first that $\delta \le 1$ . Then $|x - 2| < 1$ forces $1 < x < 3$ , hence $3 < x+2 < 5$ , hence $|x+2| < 5$ . Under that restriction,

|x^2 - 4| = |x-2|\,|x+2| < 5\,|x-2|,

which is below $\varepsilon$ as soon as $|x-2| < \varepsilon/5$ . Both requirements are met by taking the smaller of the two.

Proof. Let $\varepsilon > 0$ be given. Choose

\delta = \min\left\{1,\ \frac{\varepsilon}{5}\right\} > 0.

Suppose $0 < |x-2| < \delta$ . Since $\delta \le 1$ we have $|x-2| < 1$ , so $1 < x < 3$ and therefore $|x+2| < 5$ . Since also $\delta \le \varepsilon/5$ ,

|x^2 - 4| = |x-2|\,|x+2| < 5\,|x-2| < 5\delta \le 5 \cdot \frac{\varepsilon}{5} = \varepsilon. \qquad \blacksquare

A numerical sanity check with $\varepsilon = 0.1$ , giving $\delta = 0.02$ :

$x$	$x^2$	$\lvert x^2 - 4 \rvert$	under $\varepsilon = 0.1$ ?
$1.98$	$3.9204$	$0.0796$	yes
$1.995$	$3.980025$	$0.019975$	yes
$2.005$	$4.020025$	$0.020025$	yes
$2.02$	$4.0804$	$0.0804$	yes

The bound $5|x-2|$ is deliberately generous — the true worst case is $0.0804$ , not $0.1$ . That slack is fine. The definition asks only that you succeed, not that you succeed sharply.

What the definition rules out

The real test of a definition is what it refuses to certify.

A limit that does not exist: $\displaystyle\lim_{x\to 0}\sin\!\left(\frac{1}{x}\right)$ . As $x$ shrinks, $1/x$ grows without bound, so the sine runs through its full cycle infinitely often in any interval around $0$ . Concretely, put

p_n = \frac{1}{2\pi n + \frac{\pi}{2}}, \qquad q_n = \frac{1}{2\pi n + \frac{3\pi}{2}}.

Both sequences tend to $0$ , but $\sin(1/p_n) = 1$ for every $n$ while $\sin(1/q_n) = -1$ for every $n$ . Now suppose some $L$ were the limit and take the sceptic’s challenge $\varepsilon = 1/2$ . Any $\delta > 0$ you name contains points of both sequences, since both tend to $0$ . At those two points the outputs are $1$ and $-1$ , which are $2$ apart, so they cannot both lie within $1/2$ of a single $L$ — an interval of radius $1/2$ has width $1$ . No $L$ survives, so the limit does not exist.

For scale: at $n = 100$ , $p_n \approx 0.0015876$ and $q_n \approx 0.0015826$ . The two points differ by about five millionths, and the function values differ by $2$ .

A limit that fails by disagreement: $\displaystyle\lim_{x\to 0}\frac{1}{x}$ . Here the one-sided behaviours point in opposite directions. For $x > 0$ the values grow without bound; for $x < 0$ they decrease without bound. At $x = 10^{-6}$ the value is $10^{6}$ ; at $x = -10^{-6}$ it is $-10^{6}$ . Since any punctured interval around $0$ contains points on both sides, no single $L$ and no $\delta$ can bring all those outputs inside a common $\varepsilon$ -window. Nor can the function be repaired by assigning a value at $0$ , which is the deeper reason division by zero is left undefined rather than given some conventional answer: no choice of value makes the function continuous, because the limit itself does not exist.

Note the difference between the two failures. The $1/x$ case fails because the function is unbounded and asymmetric; the $\sin(1/x)$ case fails while the function stays neatly inside $[-1,1]$ the whole time. Boundedness is not enough.

Quantifier order, and where students actually lose

The definition begins $\forall \varepsilon\, \exists \delta$ . Swapping to $\exists \delta\, \forall \varepsilon$ produces a completely different and nearly useless statement: there is one fixed $\delta$ such that for all $\varepsilon > 0$ , the implication holds. But if $|f(x) - L| < \varepsilon$ for every positive $\varepsilon$ , then $|f(x)-L| = 0$ , so $f$ would have to be exactly equal to $L$ throughout the punctured interval. Under the swapped order, the only functions with limits would be the locally constant ones. Order is not decoration.

There is a second dependency that matters just as much, and it is where the subject turns from calculus into analysis. In the definition, $\delta$ may depend on $\varepsilon$ and on the point $a$ . Allowing that second dependency gives continuity; forbidding it gives uniform continuity:

\forall \varepsilon > 0\ \exists \delta > 0\ \forall x, y \in I: \quad |x-y| < \delta \implies |f(x)-f(y)| < \varepsilon .

Here $\delta$ must work everywhere on $I$ at once.

The canonical separating example is $f(x) = 1/x$ on the open interval $(0,1)$ . It is continuous at every point: at a point $a$ one can check that $\delta = \min\{a/2,\ \varepsilon a^2/2\}$ works, since $|x - a| < a/2$ forces $x > a/2$ and then

\left|\frac{1}{x} - \frac{1}{a}\right| = \frac{|x-a|}{ax} < \frac{2|x-a|}{a^2}.

But look at that $\delta$ : it contains $a^2$ , so it collapses as $a$ approaches $0$ . And the collapse is unavoidable. Take $\varepsilon = 1$ and consider

x_n = \frac{1}{n}, \qquad y_n = \frac{1}{n+1}, \qquad |x_n - y_n| = \frac{1}{n(n+1)} \to 0,

while $|f(x_n) - f(y_n)| = |n - (n+1)| = 1$ for every $n$ . The inputs get arbitrarily close; the outputs stay exactly $1$ apart. No single $\delta$ can serve the whole interval, so $1/x$ is continuous but not uniformly continuous on $(0,1)$ . On a closed bounded interval such as $[1,2]$ the distinction vanishes — that is the content of the Heine–Cantor theorem — which is why the issue is invisible in a first calculus course and unavoidable in a first analysis course.

Berkeley’s revenge: infinitesimals made rigorous

There is an honest postscript. The epsilon delta definition wins by refusing to talk about infinitesimals, and for a century that looked like the only option, because no real number is greater than $0$ and smaller than every positive real.

In 1966 Abraham Robinson published Non-standard Analysis, developing ideas he had announced a few years earlier, and showed that the objection was about the real numbers specifically rather than about logic. Using tools from model theory he constructed the hyperreal numbers $^{*}\mathbb{R}$ : an ordered field that properly contains $\mathbb{R}$ and genuinely contains infinitesimals — elements $\epsilon$ with $0 < \epsilon < r$ for every positive real $r$ — together with their reciprocals, which are infinite. A transfer principle guarantees that every first-order statement true of $\mathbb{R}$ remains true of $^{*}\mathbb{R}$ .

In that setting Newton’s computation becomes legitimate rather than embarrassing. With $\epsilon$ a nonzero infinitesimal,

\frac{(x+\epsilon)^2 - x^2}{\epsilon} = 2x + \epsilon,

and one does not discard the $\epsilon$ ; one takes the standard part, the unique real number infinitely close to $2x + \epsilon$ , which is $2x$ . The division is valid because $\epsilon \ne 0$ , and the final step is a well-defined operation rather than a silent deletion. The ghost has been given a body.

So why is epsilon delta still the standard? Because it costs nothing extra. It needs only the ordered field of real numbers and first-order logic, both of which a student already has. Robinson’s framework needs an ultrafilter or an equivalent non-constructive device to build $^{*}\mathbb{R}$ in the first place, which is a substantial piece of machinery to import before the first derivative. Non-standard analysis is a legitimate and occasionally more efficient foundation — it has produced genuine theorems, particularly in stochastic analysis — but it is a second language, not a replacement.

What the definition actually buys you

It is easy to treat epsilon delta as a hazing ritual and then never use it again, since in practice one computes limits with algebra and limit laws. But those limit laws have to be proved, and each proof is an epsilon delta argument. The sum rule follows from the triangle inequality with $\varepsilon/2$ for each term; the product rule needs a bounding trick like the one in the $x^2$ proof; the squeeze theorem is almost immediate once the definition is in place.

More importantly, the definition is what makes the rest of analysis expressible. Continuity is a limit statement. Differentiability is a limit of difference quotients. The Riemann integral is a limit of sums. Uniform convergence of a series of functions is the same $\forall \varepsilon\, \exists N$ pattern with the index moved. Each of these is a small variation on one sentence, and they compose because they share a grammar.

Berkeley asked what, exactly, a vanishing increment is. The answer modern mathematics settled on is that there is no such object, and none is required. There are only the two numbers $\varepsilon$ and $\delta$ , and a promise you can keep no matter which one the sceptic names first.

Frequently asked

What is the epsilon delta definition of a limit?

The limit of f(x) as x approaches a equals L if, for every epsilon greater than 0, there exists a delta greater than 0 such that whenever 0 is less than the absolute value of x minus a, which is less than delta, the absolute value of f(x) minus L is less than epsilon. In words, any tolerance you demand on the output can be met by restricting the input to a small enough punctured interval around a.

Why do we need the formal definition of a limit?

Because the intuitive version, saying f(x) gets close to L as x gets close to a, uses undefined words and hides a contradiction. Early calculus divided by an infinitesimal quantity and then discarded it as zero in the same computation. Berkeley attacked exactly this in 1734. The epsilon delta definition replaces motion and infinitesimals with inequalities between ordinary real numbers, so no quantity has to be nonzero and zero at once.

What does it mean for a limit to exist?

It means there is a single number L that works for every epsilon. If two different sequences approaching a drive f towards two different values, no single L can satisfy the definition for small epsilon, and the limit does not exist. That is why the limit of sin of 1 over x as x tends to 0 fails, and why 1 over x has no limit at 0 even though each one-sided behaviour is describable.

How do you do an epsilon delta proof?

Work backwards first. Start from the target inequality, the absolute value of f(x) minus L is less than epsilon, and manipulate it until it is expressed in terms of the absolute value of x minus a. That scratch work reveals the required delta. Then write the proof forwards: let epsilon be given, define delta explicitly, assume the input condition, and derive the output condition. For nonlinear functions you usually restrict delta to at most 1 first in order to bound the extra factor.

What is an infinitesimal?

A quantity greater than 0 but smaller than every positive real number. No such real number exists, which is why the epsilon delta definition avoids the idea entirely. Abraham Robinson made infinitesimals rigorous in 1966 by constructing the hyperreal numbers, an ordered field extending the reals that genuinely contains them, so the older intuition was eventually vindicated rather than refuted.