High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize AP Calculus (AB) Textbook > Limits & Continuity

The Epsilon Delta Definition of a Limit

The Intermediate Value TheoremPrevious Section
Exam-Like Practice QuestionsNext Section
Popular Courses
Find My Course
0:00 / 0:00

The Epsilon Delta(ϵ−δ)\color{blue}(\epsilon - \delta) Definition of a Limit

Let f(x)f(x) be a real valued function on a suitable domain DD and let a,L∈Ra,L \in \mathbb{R}. We say
lim⁡x→af(x)=Lif ∀ϵ>0,∃ δ>0 such that ∀x∈D∣x−a∣<δ⇒∣f(x)−L∣<ϵ\begin{array}{} \displaystyle\lim_{x\rightarrow a}f(x)=L \\ \\ \text{if }\forall\epsilon>0,\exists\ \delta>0\text{ such that }\forall x\in D \\ \\ |x-a|<\delta\Rightarrow|f(x)-L|<\epsilon \end{array}



0:00 / 0:00

Example: Epsilon Delta Definition of Limits

Compute the following limit using the Epsilon Delta definition

lim⁡x→4(x+2)\displaystyle \lim_{x\rightarrow4}(x+2)

Notice simply computing the limit using basic limit laws lim⁡x→4(x+2)=4+2=6\displaystyle \lim_{x\rightarrow4}(x+2)=4+2=6

We need to find a delta such that for any epsilon ∣(x+2)−6∣<ϵ|(x+2)-6|<\epsilon

Then ∣x−4∣<ϵ|x-4|<\epsilon . This happens to be the situation we need since ∣x−4∣<δ|x-4|<\delta

Thus ∀ϵ>0\forall\epsilon>0, let δ=ϵ\delta=\epsilon . Then

∣x−4∣<δ  ⟹  ∣x−4∣<ϵ  ⟹  ∣(x+2)−6∣<ϵ  ⟹  ∣f(x)−6∣<ϵ|x-4|<\delta \implies |x-4|<\epsilon\implies|(x+2)-6|<\epsilon\implies|f(x)-6|<\epsilon

∴lim⁡x→4(x+2)=6\therefore\displaystyle \lim_{x\rightarrow4}(x+2)=6
checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
Using the ϵ−δ\epsilon-\delta method, prove

lim⁡x→−24x+3=4\displaystyle\lim_{x\rightarrow-2}\frac{4}{x+3}=4