Epsilon Delta Proofs

\definition[for all $\epsilon > 0$, there exists a $\delta$ such that $if\ 0<|x-a|<\delta,\ then\ 0<|f(x)-L|<\epsilon$]{Limit Definition}{\(\lim_{x\to a} f(x) = L\)}

Let's prove \(\lim_{x\to 2} x^2 = 4\) together!

The crux of the proof is to come up with a value \(\delta\) that is a function of \(\epsilon\) assuming that \(0 < \epsilon\) that meets \(0<|x-a|<\delta\).

Oh, here's some symbols

Symbol	Definition
\(\forall\)	For all
\(\exists\)	There exisits
\(s.t.\)	Such that

And so, the formal and pretentious definition of a limit:

\(\lim_{x\to a} f(x) = L\ where\ \forall \epsilon > 0, \exists \delta > 0,\ s.t.\ 0<|x-a|<\delta \to |f(x) -L|<\epsilon.\)

This needs to go before every Epsilon Delta proof.

• Step 1: Re-write the Definition Above w.r.t. the function
• Step 2: Do scratch work to identify delta 0* Step 3: Plug it in to verify