Choosing Delta From Epsilon

Turn the Output Requirement into an Input Bound

The epsilon-delta definition asks us to respond to a specified output tolerance \(\varepsilon>0\) with an input tolerance \(\delta>0\). The previous tutorial focused on reading the quantifiers in this definition. Here we focus on the practical next step: finding a delta that makes the required output inequality follow from an estimate.

A typical argument has two ingredients. First, an algebraic estimate relates \(|f(x)-f(a)|\) to \(|x-a|\). Second, a restriction on \(|x-a|\) may be needed before that estimate is useful. We choose \(\delta\) small enough to satisfy both requirements. The minimum of two positive bounds is a convenient way to do this.

The choice need not be the largest possible one, or the only one that works. It only needs to be positive and make the proof go through. When an estimate has a factor that depends on \(x\), a common first step is to restrict \(x\) to a fixed neighborhood of \(a\), so that factor has a uniform bound.

A General Estimate for Choosing Delta

The next result packages a useful form of this reasoning. It applies when the output error is bounded by a sum of a linear and a quadratic term in the input error. The restriction \(|x-a|<1\) lets us control the quadratic term by the linear one.

Proof. Suppose first that \(A+B>0\), and choose \(\delta\) as stated. This is positive because each of \(r\), \(1\), and \(\varepsilon/(A+B)\) is positive. Let \(x\in E\) satisfy \(|x-a|<\delta\). Since \(\delta\leq r\), the assumed estimate applies. Also, \(|x-a|<\delta\leq1\), so \(|x-a|^2\leq|x-a|\). Therefore \[ |f(x)-f(a)| \leq A|x-a|+B|x-a|^2 \leq (A+B)|x-a| <(A+B)\delta \leq\varepsilon. \] The strict inequality follows from \(|x-a|<\delta\) and \(A+B>0\). Thus the chosen delta works.

If \(A+B=0\), the conditions \(A,B\geq0\) imply \(A=B=0\). For every \(x\in E\) with \(|x-a|<r\), the assumed estimate gives \(|f(x)-f(a)|\leq0\), hence \(|f(x)-f(a)|=0<\varepsilon\). Choosing \(\delta=r\) is therefore sufficient. This proves both cases. \(\square\)

A useful feature of this theorem is that it separates the work into an estimate and a choice. The estimate supplies the constants \(A\), \(B\), and the neighborhood radius \(r\); the theorem then converts them into a delta. In a particular proof, the estimate can be sharper than the one used here, but a simple bound is often easier to find and is usually enough.

Choosing Delta for the Square Function

For \(f(x)=x^2\), factor the difference of squares: \[ |x^2-a^2|=|x-a||x+a|. \] The factor \(|x+a|\) depends on \(x\), so we first restrict \(x\) to lie within distance \(1\) of \(a\). The triangle inequality then bounds that factor using only \(a\).

Proof. Fix \(\varepsilon>0\) and choose the displayed \(\delta\), which is positive. Suppose \(|x-a|<\delta\). Since \(\delta\leq1\), the triangle inequality gives \[ |x+a|=|(x-a)+2a|\leq|x-a|+2|a|<1+2|a|. \] Thus \[ |x^2-a^2|=|x-a||x+a| \leq(2|a|+1)|x-a| <(2|a|+1)\delta \leq\varepsilon. \] The first inequality is non-strict, so it remains valid when \(x=a\); in that case both sides are zero. The next inequality is strict because \(|x-a|<\delta\) and \(2|a|+1>0\). This proves the epsilon-delta condition at \(a\). \(\square\)

Keeping a Reciprocal Away from Its Singularity

For \(f(x)=1/x\), the difference can be written as \[ \left|\frac1x-\frac1a\right|=\frac{|x-a|}{|x||a|}. \] This expression is defined only when \(x\neq0\), and its denominator can become small near zero. When \(a\neq0\), choose \(\delta\leq|a|/2\). Then every \(x\) within that distance of \(a\) stays at least a fixed positive distance from zero. A second bound on \(\delta\) makes the output difference smaller than \(\varepsilon\).

Proof. Fix \(\varepsilon>0\) and choose this positive \(\delta\). Suppose \(x\in\mathbb{R}\setminus\{0\}\) and \(|x-a|<\delta\). Since \(\delta\leq|a|/2\), the reverse triangle inequality gives \[ |x|\geq |a|-|x-a|>\frac{|a|}{2}. \] In particular, \(x\) stays away from zero. Using the difference formula, \[ \left|\frac1x-\frac1a\right| =\frac{|x-a|}{|x||a|} \leq\frac{2|x-a|}{a^2} <\frac{2\delta}{a^2} \leq\varepsilon. \] The non-strict inequality is valid also when \(x=a\), when both sides are zero. The following inequality is strict because \(|x-a|<\delta\). This proves continuity at \(a\). \(\square\)

Common Pitfalls When Selecting Delta

The examples illustrate two separate jobs. A neighborhood restriction controls factors involving \(x\), such as \(|x+a|\) or \(1/|x|\). An error restriction makes the resulting bound smaller than \(\varepsilon\). A choice such as \(\delta=\min\{r,s\}\) handles both jobs at once. Omitting the neighborhood restriction can leave a factor uncontrolled; omitting the error restriction can leave the output difference too large.

A second pitfall is turning a non-strict estimate into an unjustified strict one. For example, the factorization for the square function gives \[ |x^2-a^2|\leq(2|a|+1)|x-a|, \] not necessarily a strict inequality at \(x=a\). The strict output bound comes in the next step, from \(|x-a|<\delta\) and the choice \((2|a|+1)\delta\leq\varepsilon\). Keeping these steps separate ensures that the proof also covers \(x=a\), which is always permitted by the epsilon-delta definition.

Finally, a delta formula is a sufficient choice, not a unique answer. If a positive \(\delta\) works, then any smaller positive delta works as well: it tests a smaller collection of input points. The goal is not to optimize the size of the neighborhood, but to make every implication in the proof valid for every domain point satisfying \(|x-a|<\delta\).