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Wize University Calculus 1 Textbook > Limits

The Epsilon Delta Definition of a Limit

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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}



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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)

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

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