Epsilon Delta Limit Intuition

Question

I am having a hard time grasping the definition of a limit. I initially learned this loose definition of a limit:

$\lim\limits_{x \to a}f(x)=L$ iff $f(x)$ approaches L when $x$ approaches $a$ from both directions.

Then i learned the epsilon delta definition, which states the following:

$\lim\limits_{x \to a}f(x)=L$ iff for any $\epsilon > 0$ there exists some $\delta>0$ where if $0<|x-a|<\delta$ then $|f(x)-L]<\epsilon$

The way i think about this definition(please correct me if I'm wrong) is that for any given range $\epsilon$ around L, we can find some range $\delta$ around x such that the function evaluated at those points will be within $\epsilon$ of L. However i don't understand how that statement is the same thing as saying:

$\lim\limits_{x \to a}f(x)=L$ iff $f(x)$ approaches L when $x$ approaches $a$ from both directions.

Answer 1

The way i think about this definition(please correct me if I'm wrong) is that for any given range ϵ around L, we can find some range δ around x such that the function evaluated at those points will be within ϵ of L

Yes. Rephrasing: given an error $\epsilon$ around $L$, you can find a safety margin $\delta$ around $a$ such that if $x$ is withing that safety interval $(a-\delta,a+\delta)$, then our error is small (i.e., $f(x) \in (L-\epsilon, L+\epsilon)$).

About approaching the limit from both directions.. we can formalize this using lateral limits:

• $\lim_{x \to a^+}f(x) = L_1$ iff for all $\epsilon > 0$ there is $\delta > 0$ such that $x \in (a,a+\delta) \implies f(x) \in (L_1 -\epsilon, L_1+\epsilon)$.

• $\lim_{x \to a^-}f(x) = L_2$ iff for all $\epsilon > 0$ there is $\delta > 0$ such that $x \in (a-\delta,a) \implies f(x) \in (L_1 -\epsilon, L_1+\epsilon)$.

The moral of the history is: there exists $L = \lim_{x \to a}f(x)$ if and only if both $L_1 = \lim_{x \to a^+}f(x)$ and $L_2 = \lim_{x \to a^-}f(x)$ exist and $L_1=L_2$.