The Epsilon-Delta Definition of a Limit

Most calculus students first hear a limit described as 'what the function approaches', and then assume the limit is whatever number the function value happens to be. That assumption fails the moment you hit a hole. The limit exists independently of whether the function is defined or continuous at the point. The epsilon-delta definition of limit makes that independence precise. It is the formal foundation behind every limit law, every squeeze theorem application, and every L'Hôpital's rule you will use.

Written out, the definition says: lim_x→a f(x) = L means that for every ε > 0 there exists a δ > 0 such that if 0 < |x − a| < δ then |f(x) − L| < ε. That sentence is the entire contract. The rest of this unpacks each clause, shows you the picture, and works two full epsilon-delta proofs, one for a linear function and one for x², so you can write your own.

The definition has three moving parts: the epsilon (ε), the delta (δ), and the two inequalities that connect them.

For Every ε > 0

ε is a tolerance on the output. You pick any positive number, 0.1, 0.01, 0.0001, and the function's output must stay within ε of L. The 'for every' part means the proof must work for arbitrarily small ε. If you can produce a δ for ε = 0.0001, you must also be able to produce one for ε = 0.0000001. There is no smallest ε.

There Exists a δ > 0

δ is a radius on the input side, measured from a. You get to choose δ based on the ε you were given. The statement 'there exists a δ > 0' means you must find one, and it may depend on ε. In the proofs below you will always solve for δ in terms of ε.

If 0 < |x − a| < δ Then |f(x) − L| < ε

The condition 0 < |x − a| < δ says x is within δ of a but never equal to a, the limit cares nothing about the value at a itself. The conclusion |f(x) − L| < ε says the output is within ε of L. The 'if … then' structure is a guarantee: pick any x inside the δ-window around a (except a itself) and the function value lands inside the ε-band around L.

Both Stewart §2.4 and Spivak ch. 5 state the same definition verbatim. Spivak uses it from the start of chapter 5; Stewart introduces it in §2.4 after building intuition in §2.2.

Draw the xy-plane. Put a point at (a, L).

Now draw two horizontal lines: one at y = L + ε and one at y = L − ε. The strip between them is the ε-band. Your job is to show that for every x within some distance δ of a, except a itself, the graph of f(x) stays inside that band.

Draw two vertical lines: one at x = a + δ and one at x = a − δ. The interval between them, excluding x = a, is the δ-window. The guarantee is that the entire graph of f in that window (except possibly at x = a) lies inside the ε-band.

If the graph ever pokes above L + ε or below L − ε for some x that is still within δ of a, then the δ you picked fails. You choose a smaller δ. The process of 'for every ε, find δ' is a challenge-response game: someone names an ε; you name a δ that works.

Stewart §2.4 includes this diagram in all editions. The same picture appears in Spivak ch. 5. It is the single most useful visual for understanding what the definition demands.

Prove: lim_x→2 (3x + 1) = 7.

This is the same problem Spivak ch. 5 asks and Stewart §2.4 works as an example.

Start with the definition. Let ε > 0 be given. We need a δ > 0 such that if 0 < |x − 2| < δ then |(3x + 1) − 7| < ε.

Simplify the expression inside the absolute value:

|(3x + 1) − 7| = |3x − 6| = 3|x − 2|.

We want 3|x − 2| < ε. That is equivalent to |x − 2| < ε/3.

Choose δ = ε/3. Now check: if 0 < |x − 2| < δ, then

|(3x + 1) − 7| = 3|x − 2| < 3δ = 3(ε/3) = ε.

The proof is done. δ depends only on ε, you shrink δ by the same factor 3 that the slope of the line magnifies the input error. For ε = 0.06, δ = 0.02. For ε = 0.0003, δ = 0.0001.

The same method works for any linear function: lim_x→a (mx + b) = ma + b with δ = ε/|m|, provided m ≠ 0. Stewart §2.4 gives this as a problem; the factor |m| in the denominator handles negative slopes.

Prove: lim_x→a x² = a².

This is the next step up in difficulty. The function is no longer linear, so the relationship between δ and ε is not a simple ratio. Both Stewart §2.4 and Spivak ch. 5 include this proof.

Let ε > 0 be given. We need δ > 0 such that if 0 < |x − a| < δ then |x² − a²| < ε.

Factor the difference: |x² − a²| = |x − a|·|x + a|.

The factor |x + a| is not constant, it depends on x. To control it we need a bound on how far x can be from a. The standard trick: decide in advance that δ ≤ 1. That means |x − a| < 1, which implies x is between a − 1 and a + 1, so |x + a| ≤ |x − a + 2a| ≤ |x − a| + 2|a| < 1 + 2|a|.

Now we have:

|x² − a²| = |x − a|·|x + a| < |x − a|·(1 + 2|a|)

We want this to be less than ε. If we also require |x − a| < ε/(1 + 2|a|), then the product is less than ε.

Choose δ = min(1, ε/(1 + 2|a|)). Check: if 0 < |x − a| < δ, then both conditions hold, δ ≤ 1 gives |x + a| < 1 + 2|a|, and δ ≤ ε/(1 + 2|a|) gives |x − a| < ε/(1 + 2|a|). Multiply:

|x² − a²| < (ε/(1 + 2|a|))·(1 + 2|a|) = ε.

The min() is what makes the proof work for any a. If a = 0, the bound 1 + 2|a| = 1 and δ = min(1, ε). If a = 10, the δ must be much smaller, ε/(21), because the function is steeper near x = 10 and a small input change produces a larger output change.

A common failure mode: forgetting the min() and choosing δ = ε/(1 + 2|a|) alone. That fails if a is 0 and ε is 2, because ε/(1) = 2, but |x − a| < 2 does not guarantee |x + a| < 1, x could be 1.5, making |x + a| = 1.5 and |x² − a²| = 2.25 > 2. The extra cap δ ≤ 1 prevents that.

On exams and homework you are sometimes asked: 'Find δ for a given ε' rather than writing a full proof. This is a stripped-down version of the epsilon-delta method that tests whether you can compute the bound.

Example: lim_x→3 (2x + 1) = 7. Given ε = 0.1, find δ.

Write the inequality you need: |(2x + 1) − 7| < 0.1 → |2x − 6| < 0.1 → 2|x − 3| < 0.1 → |x − 3| < 0.05. So δ = 0.05 works. Any smaller positive δ also works, 0.04, 0.01, 0.001. The question typically wants the largest δ that works, or any δ that works. When in doubt, give δ = ε/|m| for linear functions.

For x², the numerical search is more delicate. Suppose lim_x→2 x² = 4 and ε = 0.1. You need |x² − 4| < 0.1, or |x − 2|·|x + 2| < 0.1. Without a bound on |x + 2| you cannot solve directly. The algebraic method from the proof above gives δ = min(1, ε/(1 + 2|a|)) = min(1, 0.1/5) = 0.02. Check: if |x − 2| < 0.02 then |x + 2| < 4.02, so |x² − 4| < 0.02·4.02 = 0.0804 < 0.1. δ = 0.02 works.

Stewart §2.4 includes several problems of this type: find δ for a given ε for linear functions and for quadratics. The method always follows the same pattern, bound the variable part, then solve for δ in terms of ε.