Epsilon-Delta Proof — Definition, Formula & Examples

An epsilon-delta proof is a rigorous way to show that a function's limit equals a specific value by demonstrating that outputs can be made arbitrarily close to that value whenever inputs are sufficiently close to the limit point.

To prove $\lim_{x \to c} f(x) = L$ , one must show that for every $\varepsilon > 0$ , there exists a $\delta > 0$ such that whenever $0 < |x - c| < \delta$ , it follows that $|f(x) - L| < \varepsilon$ . The proof proceeds by starting with an arbitrary $\varepsilon$ , then constructing or identifying a suitable $\delta$ (typically depending on $\varepsilon$ ) that satisfies this implication.

Key Formula

\forall\, \varepsilon > 0,\; \exists\, \delta > 0 \;\text{ such that }\; 0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon

Where:

$\varepsilon$ = Any positive number representing the desired closeness of f(x) to L
$\delta$ = A positive number representing the required closeness of x to c
$c$ = The point at which the limit is evaluated
$L$ = The proposed limit value
$f(x)$ = The function being analyzed

How It Works

You begin by assuming someone gives you an arbitrary positive number

\varepsilon

, representing how close the output must be to

L

. Your job is to find a

\delta > 0

so that any

x

within distance

\delta

c

(but not equal to

c

) produces

f(x)

within distance

\varepsilon

L

. In practice, you work backward: start with

|f(x) - L| < \varepsilon

and algebraically manipulate it to find a bound on

|x - c|

. That bound gives you

\delta

. Finally, you write the proof forward—assume

0 < |x - c| < \delta

and show

|f(x) - L| < \varepsilon

follows.

Worked Example

Problem: Prove that the limit of f(x) = 3x + 1 as x approaches 2 is 7 using an epsilon-delta proof.

Set up the target inequality: We need |f(x) - L| < ε, which becomes |(3x + 1) - 7| < ε.

|3x - 6| < \varepsilon \implies 3|x - 2| < \varepsilon

Solve for the bound on |x - c|: Dividing both sides by 3 gives |x - 2| < ε/3. This tells us to choose δ = ε/3.

\delta = \frac{\varepsilon}{3}

Write the proof forward: Let ε > 0 be given. Choose δ = ε/3. If 0 < |x - 2| < δ, then:

|f(x) - 7| = |3x - 6| = 3|x - 2| < 3\delta = 3 \cdot \frac{\varepsilon}{3} = \varepsilon

Answer: Since |f(x) - 7| < ε whenever 0 < |x - 2| < δ with δ = ε/3, we have proven that lim(x→2) (3x + 1) = 7.

Why It Matters

Epsilon-delta proofs are the foundation of rigorous calculus and real analysis. Without them, statements about limits, continuity, and derivatives rest on intuition rather than proof. Mastering this technique is essential for any course in real analysis and for fields like theoretical physics or mathematical economics where precise reasoning about limits is required.

Common Mistakes

Mistake: Choosing δ before ε is given, or choosing a fixed numerical δ that does not depend on ε.

Correction: The value of δ must work for every possible ε > 0. Typically δ is expressed as a function of ε (like δ = ε/3). Always start with an arbitrary ε and derive δ from it.