Wize AP Calculus (BC) Textbook > Limits & Continuity

The Epsilon Delta Definition of a Limit

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The Epsilon Delta(ϵ−δ)\color{blue}(\epsilon - \delta)(ϵ−δ)  Definition of a Limit
Let f(x)f(x)f(x) be a real valued function on a suitable domain DDD and let a,L∈Ra,L \in \mathbb{R}a,L∈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}x→alim​f(x)=Lif ∀ϵ>0,∃ δ>0 such that ∀x∈D∣x−a∣<δ⇒∣f(x)−L∣<ϵ​

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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)x→4lim​(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=6x→4lim​(x+2)=4+2=6 

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

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

Thus ∀ϵ>0\forall\epsilon>0∀ϵ>0, let δ=ϵ\delta=\epsilonδ=ϵ . 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∣x−4∣<δ⟹∣x−4∣<ϵ⟹∣(x+2)−6∣<ϵ⟹∣f(x)−6∣<ϵ

∴lim⁡x→4(x+2)=6\therefore\displaystyle \lim_{x\rightarrow4}(x+2)=6∴x→4lim​(x+2)=6

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Using the ϵ−δ\epsilon-\deltaϵ−δ method, prove

lim⁡x→−24x+3=4\displaystyle\lim_{x\rightarrow-2}\frac{4}{x+3}=4x→−2lim​x+34​=4

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