The Epsilon-Delta Definition

From Nearby Behavior to a Precise Test

Continuity at a point is about controlling changes in function values by restricting how far the input moves. The previous tutorial used this idea to establish local consequences of continuity. Here we make the definition itself the focus: for each desired tolerance in the output, we must find an input tolerance that works for every relevant domain point.

Let \(E\subseteq\mathbb{R}\), let \(f:E\to\mathbb{R}\), and let \(a\in E\). The statement “\(f\) is continuous at \(a\)” has two nested choices. First, an arbitrary positive output tolerance \(\varepsilon\) is given. Then we must find a positive input tolerance \(\delta\) that guarantees the required bound for all \(x\in E\) sufficiently close to \(a\).

The phrase “for every \(x\in E\)” is part of the definition. We check inputs in the domain, not points outside it. The conclusion also includes \(x=a\), but that case always satisfies the output inequality because \(|f(a)-f(a)|=0<\varepsilon\). There is no need to remove \(a\) from the definition.

The order of the quantifiers matters. We choose \(\delta\) after \(\varepsilon\) is specified, and then that one \(\delta\) must work for all domain points within distance \(\delta\) of \(a\). We cannot choose a different \(\delta\) for each \(x\). Nor does the definition require one \(\delta\) to work for every possible \(\varepsilon\).

How to Choose a Delta

A proof usually begins by rewriting the output difference, \(|f(x)-f(a)|\), in terms of the input difference, \(|x-a|\). If we can bound the output difference by an expression that becomes small when \(|x-a|\) becomes small, we can use that bound to choose \(\delta\). The choice is allowed to be smaller than necessary.

Proof. Let \(\varepsilon>0\) be given. Choose \[ \delta=\min\left\{r,\frac{\varepsilon}{C}\right\}, \] which is positive because \(r\), \(\varepsilon\), and \(C\) are positive. Suppose \(x\in E\) and \(|x-a|<\delta\). Since \(\delta\leq r\), we have \(|x-a|<r\), so the assumed estimate applies. Also, since \(\delta\leq\varepsilon/C\), \[ |f(x)-f(a)|\leq C|x-a|<C\delta\leq\varepsilon. \] Thus every \(x\in E\) with \(|x-a|<\delta\) satisfies \(|f(x)-f(a)|<\varepsilon\). This is the epsilon-delta definition of continuity at \(a\). \(\square\)

This theorem is a practical sufficient test: find a local bound proportional to the input distance, then divide the desired output tolerance by the proportionality constant. It does not assert that every continuous function has such a bound with a fixed \(C\). It says that when the bound is available, it directly supplies a valid choice of \(\delta\).

Controlling a Polynomial Error

For a polynomial, factoring the difference often exposes the input distance as one factor. The other factor may still depend on \(x\), so we first restrict \(x\) to a fixed neighborhood where that factor is bounded. This two-part strategy—bound the extra factor locally, then control \(|x-a|\)—is common in epsilon-delta proofs.

Notice the role of strict and non-strict inequalities in this argument. The factor estimate gives \(|x^2-4|\leq 5|x-2|\). The desired strict conclusion follows afterward from \(|x-2|<\delta\) and \(5\delta\leq\varepsilon\). Replacing the factor estimate with a strict inequality would make it false at \(x=2\), even though the continuity proof itself would still need to include that point.

Keeping a Denominator Away from Zero

For a reciprocal function, the output difference can be factored, but its denominator must be controlled. Choosing a neighborhood that keeps the input away from zero supplies that control. This illustrates why a delta may need to satisfy more than one restriction at once.

A Further Example on a Restricted Domain

The definition works without change when the domain is not all of \(\mathbb{R}\). The implication is checked only for those \(x\) that belong to \(E\). In the next example, the formula and estimate are valid on the nonnegative domain, and the point under consideration is an interior point of that domain.

What the Definition Does and Does Not Require

Proof. Suppose \(\delta_0\) works, and let \(0<\delta\leq\delta_0\). If \(x\in E\) and \(|x-a|<\delta\), then \(|x-a|<\delta_0\). The implication for \(\delta_0\) therefore gives \(|f(x)-f(a)|<\varepsilon\). Hence \(\delta\) works as well. \(\square\)

This simple fact explains why proofs often choose a minimum of several positive quantities. Each quantity enforces one needed restriction, and their minimum enforces all of them. It also shows that a valid delta need not be the largest possible one.

A common mistake is to treat \(\delta\) as fixed independently of \(\varepsilon\). In general, a tighter output tolerance requires a smaller input neighborhood. Another is to show the desired inequality only for a particular \(x\); the definition requires one choice of \(\delta\) to work simultaneously for every \(x\in E\) within that neighborhood. Finally, an estimate involving \(|x-a|\) should be checked at \(x=a\) if its stated hypotheses include that point. A non-strict estimate such as \(|f(x)-f(a)|\leq C|x-a|\) naturally handles the zero-distance case.

At an accumulation point of the domain, continuity at \(a\) is related to the limit of \(f(x)\) as \(x\) approaches \(a\), as established in “Continuity as a Limit.” The epsilon-delta definition here states continuity directly and includes the value at \(a\). If \(a\) is isolated in \(E\), there may be a neighborhood containing no other domain points; then the condition is satisfied because the only point to check can be \(a\) itself.