MAT148H1 Chap.4 The Epsilon-Delta Definition of a Limit
The Epsilon-Delta Definition of a Limit
Reading the definition as a challenge
The statement lim f(x) = L as x → a means: for every ε > 0 there exists δ > 0 such that for all x, if 0 < |x − a| < δ then |f(x) − L| < ε. The absolute values are distances, so |x − a| < δ says x sits within δ of a, and |f(x) − L| < ε says the output sits within ε of L.
Think of a challenger who names any tolerance ε, however tiny, and your job is to answer with a window δ that guarantees every input in the window, except a itself, produces an output inside the tolerance. Because δ is chosen after ε, a correct answer is a formula in ε rather than a single number.
The function only has to be defined on an open interval around a with a possibly removed, so the value f(a), or the lack of one, never affects the limit.
From rough work to a written proof
Every ε-δ proof has two layers.
The rough work starts from |f(x) − L|, factors out |x − a|, and bounds the remaining factor by a constant k after a preliminary restriction such as |x − a| < 1. For example, near 2 the expression x² + x − 6 factors as (x − 2)(x + 3); the restriction |x − 2| < 1 places x between 1 and 3, so |x + 3| < 6. Choosing δ = min(1, ε/6) then makes both restrictions hold simultaneously.
The written proof reverses the order: fix ε, state δ explicitly, take an arbitrary x in the punctured window, and chain inequalities until the error is below ε. A good proof is self-contained, defines every variable before use, and never assumes its conclusion.
For rational functions an extra step bounds the denominator from below so that its reciprocal stays bounded above.
Showing a limit fails
Negating the definition swaps each quantifier: lim f(x) ≠ L exactly when there exists ε > 0 such that for every δ > 0 there is some x with 0 < |x − a| < δ and |f(x) − L| ≥ ε. In plain terms, one fixed tolerance defeats every window.
The usual disproof names a specific ε, lets δ be arbitrary, and builds a witness point from δ, often something like a + min(δ, 1)/2 so that it lies strictly inside the window.
To show that no limit exists at all, a universal quantifier over L goes in front, and ε may then depend on L.
Why the definition matters for the rest of the course
Every limit law, the uniqueness of limits, continuity, and the derivative itself rest on this sentence.
Uniqueness, for instance, follows by taking ε as half the distance between two candidate limits: the two ε bands are disjoint, and the graph near a cannot lie in both. Equivalent reformulations, such as replacing every real M by every integer M in the definition of an infinite limit, are proved the same way, by fixing the arbitrary object first and choosing the window second.
Continuity at a uses the identical sentence with L replaced by f(a), so this proof shape carries straight into the continuity arguments that follow. A practical habit is to test any proposed δ numerically: with ε = 0.6 the rule above gives δ = 0.1, and x = 2.09 produces an error near 0.458, comfortably inside the tolerance.
What this chapter covers
- 01
Quantifier order in the ε-δ definition
- 02
Epsilon bands and delta windows on a graph
- 03
Factor and bound routine for polynomials
- 04
Choosing δ = min(1, ε/k)
- 05
Negation of the limit definition
- 06
Witness points for limits that fail
- 07
Uniqueness of limits
Worked example · free
An ε-δ proof that x² + x tends to 6 at 2
- 2Rough work: |x² + x − 6| = |x − 2| |x + 3|. If |x − 2| < 1 then 1 < x < 3, so 4 < x + 3 < 6 and |x + 3| < 6.
- 1Let ε > 0 and choose δ = min(1, ε/6), so δ ≤ 1 and δ ≤ ε/6.
- 1Let x satisfy 0 < |x − 2| < δ. Since δ ≤ 1, we have 1 < x < 3 and therefore |x + 3| < 6.
- 2Then |x² + x − 6| = |x − 2| |x + 3| < (ε/6)(6) = ε. Since ε was arbitrary, the limit equals 6.
Key terms
- Epsilon Band
- An epsilon band is the horizontal strip of points whose height lies strictly between L − ε and L + ε. A limit statement requires the graph near a to stay inside this strip for every positive ε.
- Delta Window
- A delta window is the set of inputs x with 0 less than |x − a| less than δ, an open interval around a with the centre removed. The prover chooses its radius after the tolerance is given.
- Punctured Neighbourhood
- A punctured neighbourhood is an open interval around a point with that point deleted. Limits only inspect inputs in such a set, so the value of the function at the point itself is irrelevant.
- Witness Point
- A witness point is a specific input, built from an arbitrary δ, lying inside the delta window yet sending the output outside the epsilon band. Producing one for every δ proves that a proposed limit fails.
- Minimum Rule
- The minimum rule is the choice δ = min(δ₁, δ₂), which ensures that two separate restrictions on |x − a| hold simultaneously, typically one bounding a factor and one shrinking the error.
The Epsilon-Delta Definition of a Limit FAQ
How do you find delta in an epsilon delta proof?
Start from |f(x) − L| and factor out |x − a|. Restrict |x − a| less than 1 to bound the leftover factor by a constant k, then take δ = min(1, ε/k). That choice makes the bound and the smallness work together in the final inequality.
Why is the condition 0 less than |x − a| included in the limit definition?
It deletes the point a from the set of tested inputs, so the limit describes behaviour near a rather than at a. Functions with holes, or with a value at a that differs from nearby values, still have well-defined limits under this convention.
How do you prove that a limit does not equal a given number?
Use the negation: pick one positive tolerance, let the window radius be arbitrary, and construct an input inside that window whose output misses the target by at least the tolerance. The input may depend on the radius, but the tolerance must be fixed first.
Can delta be any number smaller than the one you found?
Yes. Shrinking the window only removes inputs from consideration, so every inequality that held for the larger radius still holds for the smaller one. Proofs report one convenient choice, and different correct proofs often state different but equally valid formulas.
Exam move
Each day, write the definition from memory with the quantifiers in order, then complete one forward proof and one negation proof. For forward proofs use a fixed checklist: factor, cap by 1, compute k, state δ = min(1, ε/k), write up. Test your δ numerically at one ε and one x near a. For negations, always state ε first and construct the witness explicitly from δ.
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