Why We Need It
The intuitive definition — "f(x) gets close to L as x gets close to a" — is too vague for rigorous mathematics. What does "close" mean? The epsilon-delta definition replaces intuition with precision.
The Definition
Think of it as a game: your opponent picks any ε (output tolerance). You respond with δ (input restriction). If you can always win, the limit exists.
Opponent: "I want f(x) within 0.01 of L." You: "Then keep x within δ = 0.005 of a." A proof shows you can always find such a δ, no matter how small ε is.
Worked Proof
Frequently Asked Questions
The ε-δ Game
The cleanest way to understand the definition is as a two-player game. Player 1 (the sceptic) picks any ε > 0, no matter how small. Player 2 (the mathematician) must respond with a δ > 0 such that the guarantee holds: whenever x is within δ of a (but not equal to a), f(x) is within ε of L. If Player 2 can always win, the limit exists.
The key phrase is "for every ε > 0". Player 2 must have a strategy that works for any ε the sceptic picks — including ε = 0.001, ε = 0.0000001, and ε = 10⁻¹⁰⁰.
Proving a Quadratic Limit
One-Sided Limits
For a one-sided limit lim(x→a⁺) f(x) = L, the definition is the same but the condition changes to 0 < x − a < δ (for the right-hand limit) or −δ < x − a < 0 (for the left). The two-sided limit exists exactly when both one-sided limits exist and are equal.
Formal Definition of Continuity
A function is continuous at a precisely when lim(x→a) f(x) = f(a). In ε-δ language: for every ε > 0, there exists δ > 0 such that |x − a| < δ (note: no "0 <" this time — we include x = a) implies |f(x) − f(a)| < ε. The only difference from the limit definition is that x = a is now allowed.
When You Actually Need This
In a first calculus course, you rarely need to write ε-δ proofs. The definition matters for three reasons: understanding why limits behave the way they do (why does lim x² = a² follow from lim x = a?), rigorous analysis courses where you must prove every theorem, and settling disputes when intuition fails (why does lim(x→0) sin(1/x) not exist?). For computations, the limit laws and L'Hôpital's rule are sufficient.
Two More Worked Proofs
Proving a Limit Does NOT Exist
To show lim(x→0) sin(1/x) does not exist, you show that for any proposed L, the ε-δ condition fails. Take ε = 1/2. No matter how small δ is, in the interval (−δ, δ) the function sin(1/x) takes every value between −1 and 1 infinitely often (because 1/x → ∞). So you can always find x within δ of 0 where |sin(1/x) − L| > 1/2, for any L.
Sequential Characterisation
An equivalent condition: lim(x→a) f(x) = L if and only if for every sequence xₙ → a (with xₙ ≠ a), we have f(xₙ) → L. This is sometimes easier to work with than ε-δ. To disprove a limit, find two sequences approaching a that give different values: for sin(1/x), the sequence xₙ = 1/(nπ) → 0 gives sin(1/xₙ) = 0, but xₙ = 2/(π(4n+1)) → 0 gives sin(1/xₙ) = 1.