Why Epsilon-Delta Proofs Need Careful Logic
A limit proof is not just a calculation that makes the output difference look small. It must establish a particular logical statement: for every requested output tolerance, there is an input tolerance that guarantees it. The order of those choices matters, as does the direction of every implication used in the argument.
This tutorial focuses on diagnosing errors in such proofs. The most common problems are choosing \(\delta\) before fixing \(\varepsilon\), treating the desired conclusion as if it were already known, and trying to control a quadratic expression without bounding its other factor.
The condition \(0<|x-a|\) excludes \(x=a\): a limit concerns values near the input, not necessarily the value at the input itself. In a proof, it is not enough to propose a formula for \(\delta\). One must verify that the formula is positive and that it makes the required implication true.
Error 1: Choosing Delta Before Epsilon
The definition begins with “for every \(\varepsilon>0\)” and only then allows a choice of \(\delta\). This means the proof must handle an arbitrary positive tolerance. A fixed \(\delta\) might work for some tolerances but fail for smaller ones.
Worked Example: A Fixed Delta Fails for a Linear Limit
Consider the claim \(\lim_{x\to 0}x=0\). A faulty argument says: “Choose \(\delta=0.1\). Then if \(0<|x|<\delta\), we have \(|x|<\varepsilon\).” The final inequality does not follow for an arbitrary \(\varepsilon\).
Test the proposed choice with \(\varepsilon=0.05\), and take \(x=0.075\). Then
Thus the proposed \(\delta\) does not satisfy the definition. The problem is not that \(0.1\) is always an unusable input tolerance; it is that it was presented as a fixed choice that should work for every positive \(\varepsilon\).
To repair the argument, let \(\varepsilon>0\) be arbitrary and choose \(\delta=\varepsilon\). If \(0<|x|<\delta\), then
Here \(\delta>0\), and the implication required by the definition has been verified. The chosen \(\delta\) explicitly responds to the given \(\varepsilon\).
A proof can state its choices in a different order for convenience, but it cannot change their logical dependence. If a proposed \(\delta\) is selected using no information about \(\varepsilon\), ask whether it has genuinely been shown to work even when \(\varepsilon\) is arbitrarily small.
Error 2: Assuming the Desired Conclusion
To prove a limit, the required direction is from a small input difference to a small output difference. In symbols, the proof starts from \(0<|x-a|<\delta\) and must establish \(|f(x)-L|<\varepsilon\). Deriving the input condition from the output condition reverses the implication. Even if the reversed statement is true, it does not prove the statement required by the definition.
Worked Example: Reversing an Implication
Consider \(f(x)=x^2\), with input \(a=2\) and proposed limit \(L=4\). For \(\varepsilon=1\), the required implication is
A faulty argument starts by assuming \(|x^2-4|<1\), factors the expression, and concludes that \(x\) is close to \(2\). But it has assumed the conclusion it was meant to prove. It has not shown that a sufficiently small input difference guarantees the output bound.
The direction matters even for a simple numerical implication. For \(x=0\), the output inequality is false because
while the input inequality \(|0-2|<3\) is true. Thus “the output bound implies the input bound” cannot be substituted for the required claim that the input bound implies the output bound.
A valid proof instead begins with the input restriction and estimates the output. For instance, set \(\delta=\min(1,\varepsilon/5)\) for arbitrary \(\varepsilon>0\). If \(0<|x-2|<\delta\), then \(|x-2|<1\), so \(|x+2|\leq |x-2|+4<5\). Therefore
The last inequality follows from \(\delta\leq\varepsilon/5\). This proof starts from the hypothesis and reaches the required conclusion, rather than assuming it.
A useful check is to label the two statements in the proof: the hypothesis is the input condition, and the conclusion is the output condition. Each estimate must carry the argument from the former to the latter. Rewriting the desired output inequality and manipulating it backwards can help suggest what to try, but it is not itself a proof unless the implication is subsequently established in the correct direction.
Error 3: Letting Delta Depend on the Input
The definition requires one positive number \(\delta\) to be chosen after \(\varepsilon\) is given. That \(\delta\) must work for every \(x\) in the specified neighborhood. A proposed choice such as \(\delta=\varepsilon/|x+a|\) is not valid: it changes with \(x\), the very input that is allowed to vary after \(\delta\) has been chosen.
This issue often appears when factoring a quadratic difference. For \(f(x)=x^2\) near \(a\), the relevant identity is
The input condition controls \(|x-a|\), but it does not immediately control \(|x+a|\). To use the factorization, the proof must first restrict \(x\) to a neighborhood where the second factor has a fixed bound.
Worked Example: A Quadratic Bound Without a Local Restriction
Consider again \(x^2\) near \(2\), with target value \(4\). A faulty proof observes the factorization and effectively treats \(|x+2|\) as though it were at most \(1\). It then claims that \(|x^2-4|\leq|x-2|\), and proposes \(\delta=\varepsilon\).
That factor estimate is not true near \(2\): for \(x=2\), \(|x+2|=4\), not at most \(1\). The proposed delta also fails directly. Take \(\varepsilon=0.1\) and \(x=2.05\). Then
Since \(0.2025\not<0.1\), this \(\delta\) does not guarantee the requested output tolerance.
To repair the proof, first require \(\delta\leq1\). Then \(|x-2|<\delta\leq1\) gives
For arbitrary \(\varepsilon>0\), choose \(\delta=\min(1,\varepsilon/5)\). If \(0<|x-2|<\delta\), then
The factor is now bounded by a fixed number, \(5\), throughout the chosen neighborhood. The delta depends on \(\varepsilon\), not on the particular \(x\), and the displayed estimate verifies the result.
Why the Local Restriction Works
The quadratic argument uses the triangle inequality, \(|u+v|\leq |u|+|v|\), to turn a bound on distance from \(a\) into a bound on the other factor. For any real \(a\), if \(|x-a|<1\), then
Thus a restriction \(\delta\leq1\) provides the fixed bound \(|x+a|<1+2|a|\). This is the key step: the factor \(|x+a|\) is not assumed small; it is shown to stay bounded when \(x\) is confined to a suitable neighborhood of \(a\).
Indeed, this choice is positive for every \(\varepsilon>0\). If \(0<|x-a|<\delta\), then \(|x+a|<1+2|a|\), and consequently
The strict inequality before the final comparison follows from \(|x-a|<\delta\); the final comparison follows from the definition of the minimum. This verifies the required implication.
A Checklist for Diagnosing a Proof
When reviewing an epsilon-delta argument, check each of the following points.
- Quantifier order: Is \(\varepsilon>0\) arbitrary before \(\delta\) is chosen?
- Positivity: Is the proposed \(\delta\) strictly positive for every allowed \(\varepsilon\)?
- Dependencies: Does \(\delta\) depend only on permitted fixed data and \(\varepsilon\), not on \(x\)?
- Implication direction: Does the reasoning start from \(0<|x-a|<\delta\) and prove \(|f(x)-L|<\varepsilon\)?
- Every factor: If the output difference is factored, has each factor been controlled by a stated estimate?
- Final verification: Do the estimates actually end with a strict bound by \(\varepsilon\)?
These checks do not replace the proof. They help locate the exact point where a proposed argument succeeds or fails. A proof is complete only when its choices are valid and the required implication has been demonstrated for every input in the stated neighborhood.
Check Your Understanding
For each question, identify the logical issue or give the required justification.
- In the definition of a limit, which quantity is chosen first: \(\varepsilon\) or \(\delta\)?
- Why does a proof of “the output difference is small implies the input difference is small” not establish a limit?
- Why is \(\delta=\varepsilon/|x+a|\) not a valid choice in a proof where \(x\) is the variable input?
- For \(x^2\) near \(a\), what restriction on \(|x-a|\) gives the bound \(|x+a|<1+2|a|\)?
- For \(x^2\) near \(2\), check the output difference when \(x=2.05\) and \(\varepsilon=0.1\). Does \(\delta=\varepsilon\) work?
