Epsilon-Delta limits, *not* for a straight line. Parabola example

Epsilon-Delta limits, *not* for a straight line. Parabola example: "

So the general tactic for straight lines:



f(x) = 2x

Show by epsilon-delta definition of limit, as lim x->2, f(x) tends to 4.

let, ε>0, and choose 0<δ<ε/2



(|x - 2|<δ)→ |f(x) - 4| = |2x - 4| = 2|x - 2| < 2δ < ε

No problem, but what about for a parabola?



g(x) = ax² for some a in R.

Show as lim x -> b, g(x) tends to ab²



let, ε>0, and choose 0<δ<?

|x - 2|<δ

|f(x) - ab²| = |ax² - ab²| = a|x²-b²| ??

I don't have a clear idea what to do here; I tried;

a|(x+b)(x-b)| = a|(x+b)||(x-b)| < a|(x+b)|δ

Is it ok to simply choose:

0<δ<ε/|a(x+b)|

after all, it's just a positive number.



I really hope this post's question can be answered in one word beginning with y.

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