How One Epsilon Can Disprove a Limit
A limit statement makes a demanding promise: for every positive output tolerance, there is an input distance that keeps all sufficiently nearby function values within that tolerance of the proposed limit. To refute the statement, it is enough to find one positive tolerance for which that promise fails, no matter how small the input distance is made.
The central idea is to keep the output tolerance fixed while allowing the input distance to be arbitrary. Then, for each proposed input neighborhood, we produce at least one permitted input whose function value remains too far from the claimed limit.
The Definition and Its Negation
Let \(f:D\to\mathbb{R}\), where \(D\) is a subset of \(\mathbb{R}\), and suppose \(a\) is an accumulation point of \(D\). The statement \(\lim_{x\to a}f(x)=L\) means that every positive output tolerance can be met by choosing a suitable positive input distance.
The input condition is punctured: \(x=a\) is excluded. A limit concerns function values at nearby inputs, not necessarily the value \(f(a)\). The output condition is strict: values at distance exactly \(\varepsilon\) from \(L\) do not satisfy \(|f(x)-L|<\varepsilon\).
To negate the definition, negate its quantifiers and the implication. The negation of “for every \(\varepsilon>0\)” is “there exists an \(\varepsilon>0\),” and the negation of “there exists \(\delta>0\)” is “for every \(\delta>0\).” For an input satisfying the punctured-neighborhood condition, failure of the strict output inequality means the distance is at least \(\varepsilon\).
The number \(\varepsilon_0\) is an epsilon witness. It is chosen once and stays fixed for every \(\delta\). The input \(x\) may depend on \(\delta\), but for each \(\delta\) it must be in the domain, differ from \(a\), and have a function value at least \(\varepsilon_0\) away from \(L\).
Why the Witness Is Enough
Suppose a proposed limit were true. Its definition would apply to the particular positive tolerance \(\varepsilon_0\), giving some \(\delta_0>0\) such that every domain input with \(0<|x-a|<\delta_0\) satisfies \(|f(x)-L|<\varepsilon_0\). But the witness property applies to this very \(\delta_0\): it supplies a domain input in that punctured neighborhood with \(|f(x)-L|\geq\varepsilon_0\). The two inequalities cannot both hold for the same input. Thus the proposed limit is false.
Worked Example: A Wrong Proposed Limit for a Quadratic
Worked Example: A Wrong Proposed Limit for a Quadratic
Consider \(f(x)=x^2\) as \(x\to 0\). Refute the proposed limit \(L=1\).
Choose the fixed witness \(\varepsilon_0=\frac12\). Let \(\delta>0\) be arbitrary. Set \(x=\min\{\frac{\delta}{2},\frac12\}\). This input is positive, so \(0<|x-0|=x\). Also \(x\leq\frac{\delta}{2}<\delta\), so it lies in the required punctured neighborhood.
Since \(x\leq\frac12\), we have \(0\leq x^2\leq\frac14\). Therefore the distance from the proposed limit is
This construction works for every \(\delta>0\) with the same \(\varepsilon_0\). Hence the proposed limit \(1\) is false. As a check, the constructed input always satisfies the two required conditions: \(0<x<\delta\), and its output distance is at least \(\frac34\), which is greater than the witness \(\frac12\).
Worked Example: A Wrong Proposed Limit for an Affine Function
Worked Example: A Wrong Proposed Limit for an Affine Function
Consider \(f(x)=2x+1\) as \(x\to 2\). Refute the proposed limit \(L=10\).
Choose \(\varepsilon_0=1\). For an arbitrary \(\delta>0\), define \(t=\min\{\frac{\delta}{2},\frac14\}\) and choose \(x=2+t\). Because \(t>0\), we have \(x\ne 2\) and \(0<|x-2|=t\). Since \(t\leq\frac{\delta}{2}<\delta\), this input is within the punctured \(\delta\)-neighborhood of \(2\).
We also have \(0<t\leq\frac14\), so \(2<x\leq\frac94\). Thus \(5<f(x)=2x+1\leq\frac{11}{2}\). In particular, \(f(x)<10\), and
For every positive \(\delta\), the chosen \(x\) is a valid domain input and its function value stays at least \(1\) away from \(10\). This verifies the witness and refutes the proposed limit. Notice that the input may approach \(2\) as \(\delta\) shrinks, but its output remains near \(5\), not \(10\).
Worked Example: Oscillating Values Near the Input
Worked Example: Oscillating Values Near the Input
For \(x\ne 0\), let \(f(x)=\sin(1/x)\). Refute the proposed limit \(L=0\) as \(x\to 0\).
Choose \(\varepsilon_0=\frac12\). Let \(\delta>0\) be arbitrary. Since \(\mathbb{N}=\{1,2,3,\dots\}\), we can choose a positive integer \(n\) large enough that \(2\pi n+\frac{\pi}{2}>\frac1\delta\). Define
The denominator is positive and exceeds \(1/\delta\), so \(0<x<\delta\). In particular, \(x\ne 0\), and \(x\) is in the domain. Substituting this input gives
Therefore \(|f(x)-0|=1\geq\frac12=\varepsilon_0\). For every \(\delta>0\), we have constructed a valid input within distance \(\delta\) of zero whose function value misses the proposed output tolerance. The proposed limit is false. The check is direct: \(0<|x|<\delta\), and the output distance is exactly \(1\).
A Reliable Witness-Building Method
When a limit claim seems false, first decide what output values would contradict the proposed limit. Then find a fixed positive distance that separates those values from \(L\). A successful construction must work no matter what positive input distance another person proposes.
Identify the input point \(a\), proposed limit \(L\), and domain \(D\).
Select \(\varepsilon_0>0\) that the problematic output values fail to meet.
Do not assume a particular neighborhood size; the construction must handle every one.
Give \(x\in D\) with \(0<|x-a|<\delta\), often by choosing a nearby point or a suitable member of a sequence.
Calculate or bound \(|f(x)-L|\) and show it is at least \(\varepsilon_0\).
Each condition matters. A point outside the domain is not a valid witness. The point \(x=a\) is excluded even if \(f(a)\) is far from \(L\). A tolerance that changes with \(\delta\) does not establish one epsilon witness. And showing only that an output is not equal to \(L\) is insufficient: its distance from \(L\) must be bounded below by the same positive number at every scale.
Common Mistake: Using the Excluded Input
Suppose someone tries to refute \(\lim_{x\to 0}x^2=1\) by taking \(x=0\), noting that \(|f(0)-1|=1\). This is not a valid witness. For every \(\delta>0\), the definition tests inputs satisfying \(0<|x-0|<\delta\), and \(x=0\) fails the strict lower bound.
The correction is to choose a nearby input that is not zero, such as \(x=\min\{\frac{\delta}{2},\frac12\}\), as in the worked example. It satisfies \(0<x<\delta\), and its output distance from \(1\) is at least \(\frac34\). The witness must meet both the input and output conditions.
Why Epsilon Witnesses Matter
A graph or a table can suggest that a proposed limit is wrong, but it only displays finitely many inputs or a visual approximation. An epsilon witness gives a precise argument at every neighborhood size. It explains exactly how the limit definition fails: a fixed output tolerance cannot be enforced, however small the input distance is chosen.
This method is also the logical counterpart of an epsilon-delta proof. A proof of a limit handles every positive epsilon by finding a suitable delta. A disproof identifies one positive epsilon for which no delta can work. The difference is not a numerical test; it is the order of the quantifiers.
Check Your Understanding
For each question, pay attention to which quantities must be fixed and which may depend on others.
- State the quantifiers and inequalities in the negation of \(\lim_{x\to a}f(x)=L\).
- In an epsilon-witness argument, which quantity is fixed before \(\delta\) is chosen?
- Why is \(x=a\) not a valid witness, even if \(|f(a)-L|\) is large?
- For \(f(x)=x^2\) as \(x\to 0\), proposed limit \(1\), explain why choosing \(x=\min\{\frac{\delta}{2},\frac12\}\) gives \(0<|x|<\delta\).
- Why is it not enough to show that \(f(x)\ne L\) for points arbitrarily close to \(a\)?
