Translating basic limit intuition to epsilon delta definition

Question

Early on in precalculus/calculus I learned that a $$\lim_{x\to c} ~ f(x)$$ does not exist if $$\lim_{x\to c^{+}} ~ f(x) \neq \lim_{x\to c^{-}} ~ f(x)$$

I'm having trouble understanding how to translate/map this concept to the $\epsilon-\delta$ definition of limits and the idea of errors($\epsilon$) and distances($\delta$). Why does the definition have those specific constraints?

Answer 1

Let $a=\lim_{x\to c^+}f(x)$, $b=\lim_{x\to c^-}f(x)$. Let $\epsilon =\frac{|a-b|}2>0$. Then there exists $\delta_+$ such that $c<x<c+\delta_+$ implies $|f(x)-a|<\epsilon$. And there exists $\delta_->0$ such that $c>x>c-\delta_-$ implies $|f(x)-b|<\epsilon$. Go figure.

Answer 2

The constraint is basically saying that for the proposition

For every $\epsilon>0$ there exists $\delta>0$ such that if $|x-c|<\delta$, then $|f(x)-l|<\epsilon$

to be true, both

For every $\epsilon>0$ there exists $\delta>0$ such that if $x-c<\delta$, then $|f(x)-l|<\epsilon$

For every $\epsilon>0$ there exists $\delta>0$ such that if $c-x<\delta$, then $|f(x)-l|<\epsilon$

need to be true. It casually happens that the two above definitions are those of the two one-sided limits.

Answer 3

$$\lim_{x\to c^{+}} ~ f(x) =l$$ if and only if: Given $\epsilon>0,\exists \delta>0$ such that if $x\in (c, c+\delta)$, the $|f(x)-l|<\epsilon$. $$lim_{x\to c^{-}} ~ f(x)=l$$ is defined accordingly by taking $x\in (c-\delta, c).$

Answer 4

The short answer to your question is because it is useful. The longer answer is that the natural understanding of a limit is "when $x$ get really close to $x_0$, what (if anything) does $f(x)$ get really close to?" The $\varepsilon-\delta$ Convention formalizes this idea in a way that is very precise and works in every case. The limit exists if there is a number $l$ such that for any amount of error $\varepsilon$, I can find a distance $\delta$ such that if the distance between $x$ and $x_0$ is less than $\delta$, the error between $f(x)$ and $l$ is less than $\varepsilon$.