The ε–δ Definition of a Limit
“A limit is the promise: 'name any error you like, and I can meet it.'”
The formula
lim(x→a) f(x) = L ⟺ ∀ε>0, ∃δ>0 : 0 < |x−a| < δ ⟹ |f(x)−L| < εHow to read it: no matter how small a target error ε you demand, there is an input window δ that keeps f(x) within ε of L
- ε
- — the target error allowed on the output — the challenge 'get this close'
- δ
- — the input window on x that meets the demand
- |x−a|
- — the distance of input x from a
- |f(x)−L|
- — the error between output f(x) and the target L
The hook
Mathematics pinned down the vague phrase 'gets arbitrarily close' as a challenge-and-response game.
In plain words
If, however small an error ε your opponent names, you can shrink the input window δ so that f(x) stays within ε of L, then the limit is L.
The intuition
See it as a game. Opponent: 'Get the output within 0.01 of L (that's ε).' You: 'Sure—just keep the input within δ of a.' If you can always answer any smaller ε with some δ, you win, and that is 'the limit is L'. Closeness becomes a rule, not a feeling.
How it's built
Two pieces. ε is the demand on the output (how close), δ is the response on the input (how narrow). The order matters—ε is given first, then δ is chosen to match. Usually δ depends on ε.
Example
Show lim(x→1)(2x+1)=3. Given ε=0.1, we need |2x+1−3|=2|x−1|<0.1, i.e. |x−1|<0.05, so δ=0.05 works. In general δ=ε/2 answers any ε, so the limit is 3.
Common mistake
You do not pick δ first and then find ε. ε always comes first as the challenge; δ follows as the response. Swap the order and the definition collapses.
Where it's used
This definition is the bedrock of all calculus—continuity, derivatives, integrals, and convergence are all made rigorous in this language. Removing the vagueness let mathematics handle infinity safely.
Where it came from
Cauchy sowed the seed in the 1800s and Weierstrass finished it with the symbols ε and δ. It turned limits, once fudged with 'infinitesimals', into a finite game of inequalities.
Prerequisites
Quick check
For lim(x→1)(2x+1)=3, you are given ε=0.2. Which δ guarantees |f(x)−3|<0.2?
- δ=0.1✓
- δ=0.2
- δ=0.4
- δ=2
Practice
For f(x)=3x, lim(x→2)=6. Find the largest δ that guarantees an error under ε=0.3.
Answer: 0.1
- need |3x−6|=3|x−2|<0.3
- → |x−2|<0.1, so δ=0.1
Takeaway: δ usually scales with ε—here δ=ε/3.
For f(x)=4x, lim(x→0)=0. What is the largest δ guaranteeing error under ε=0.8?
Answer: 0.2
- need 4|x|<0.8
- → |x|<0.2, so δ=0.2
Takeaway: A steeper slope needs a narrower δ for the same ε.
For f(x)=5x, lim(x→3)=15. Using the rule δ=ε/5, what δ meets ε=1?
Answer: 0.2
- δ=ε/5 = 1/5
- = 0.2
Takeaway: For a line you can always answer with δ=ε/(slope).
Explain in your notebook why 'for every ε there exists a δ' exactly captures 'gets arbitrarily close'.
Answer: undefined
- 'arbitrarily close' means 'reach any target error, no matter how small'
- having a δ for every ε means every demand can be answered
- so no error is out of reach = it truly approaches
Takeaway: 'every ε' captures 'arbitrarily', 'a δ exists' captures 'reachable'.