Epsilon-Delta Proofs for Constant Functions

Why Constant Functions Are a First Epsilon-Delta Proof

A constant function has the same output at every input. Its graph is a horizontal line, and its values do not change as the input approaches a chosen point. That makes it an ideal first example for using the epsilon-delta definition: the output error can be calculated exactly, without needing to estimate it.

The goal is not merely to observe that the function stays fixed. We must match the order of the definition: start with any positive output tolerance \(\varepsilon\), then give a positive input radius \(\delta\) that guarantees the required output error whenever the input is sufficiently close to the point.

The Epsilon-Delta Condition

For a function \(f:\mathbb{R}\to\mathbb{R}\), the statement \(\lim_{x\to a} f(x)=L\) means that for every \(\varepsilon>0\), there exists a \(\delta>0\) such that, for every \(x\) in the domain,

The condition \(0<|x-a|\) says that \(x\) is not the input point \(a\) itself. The condition \(|x-a|<\delta\) says that \(x\) lies within distance \(\delta\) of \(a\). The conclusion measures the output's distance from the proposed limit \(L\).

The quantifiers matter. The tolerance \(\varepsilon\) is given first and may be any positive number, however small. We must then choose a positive \(\delta\). After that choice, the implication must hold for every \(x\) satisfying the input condition. The choice of \(\delta\) cannot depend on which such \(x\) is selected.

The Constant-Function Result

Let \(c\) be a fixed real number, and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=c\) for every real input \(x\). At any real number \(a\), the claimed limit is \(c\). To check the output error, substitute the constant function and proposed limit directly into the definition:

This error is zero for every \(x\), so it is smaller than every positive \(\varepsilon\). The input restriction does not need to do any work: as soon as \(\delta\) is positive, any \(x\) that satisfies \(0<|x-a|<\delta\) has output error zero.

Complete Epsilon-Delta Proof

We now verify the theorem in the precise order required by the definition. Fix any real numbers \(a\) and \(c\), and let \(f(x)=c\) for all \(x\in\mathbb{R}\). We will prove that \(\lim_{x\to a}f(x)=c\).

Thus, for every \(\varepsilon>0\), the positive choice \(\delta=1\) ensures that every \(x\) with \(0<|x-a|<\delta\) satisfies \(|f(x)-c|<\varepsilon\). By the epsilon-delta definition, \(\lim_{x\to a}f(x)=c\). The numbers \(a\) and \(c\) were arbitrary, so the result holds at every input point and for every constant value.

Why the Choice Works

The proof works because the quantity controlled by the definition is the output error \(|f(x)-L|\). For a constant function with proposed limit equal to its constant value, that expression simplifies to zero before any condition on \(x\) is used. Since every allowed tolerance is positive, zero is strictly smaller than the tolerance.

This also explains why the chosen \(\delta\) does not need to be calculated from \(\varepsilon\). In other limit proofs, the output error may vary with \(x\), so the input radius must be chosen to control that variation. Here there is no variation to control: the output error is always zero. Choosing \(\delta=1\) is convenient, but not special. For any fixed \(\delta>0\), the same calculation proves the implication.

The strict inequality at the end is important. The definition asks for \(|f(x)-L|<\varepsilon\), not merely \(|f(x)-L|\leq\varepsilon\). Because \(\varepsilon\) is strictly positive, the equality \(|f(x)-c|=0\) gives the required strict inequality.

Worked Examples

Common Mistakes and Their Corrections

A tempting but invalid choice is \(\delta=0\), on the grounds that the output error is zero. For example, one might write:

This fails because the definition requires \(\delta>0\). Moreover, there is no \(x\) satisfying \(0<|x-a|<0\), so the implication would have no permitted inputs to check. A valid proof must exhibit a positive radius, such as \(\delta=1\), and then verify the output inequality for every input within that radius, except \(a\).

Another error is to confuse a constant function with the identity function and claim that its output error equals the input distance. For instance, the calculation

does not follow when \(f(x)=c\). The correct substitution is \(f(x)=c\), which gives \(|f(x)-c|=|c-c|=0\). The input distance \(|x-a|\) appears in the hypothesis of the definition; it does not equal the output error unless a separate calculation establishes that equality.

Why This Proof Matters

This example establishes a useful pattern for direct proofs: identify the output error in the conclusion, substitute the function and proposed limit, and then choose a positive input radius that makes the required implication true. In the constant case, simplification makes the error exactly zero, so the choice of radius is immediate.

The proof also illustrates how a statement about nearby behavior differs from simply evaluating a function at a point. The definition asks for a condition involving all sufficiently nearby inputs, not just the value at \(a\). For a constant function, every nearby input has the same output, so the output remains within any positive tolerance of the constant value.

When proving this result, keep the logical order visible: \(\varepsilon>0\) is arbitrary; then \(\delta>0\) is chosen; then an arbitrary \(x\) satisfying \(0<|x-a|<\delta\) is considered; finally the output error is shown to be less than \(\varepsilon\). Once each part is stated, the proof is short, but every part is needed to meet the definition.