Question

Prove \[\lim_{x \rightarrow 0} \sqrt{4-x} = 2\]using \(\epsilon\) - \(\delta\) definition.

Answer 1

For δ one:
0 < x-0 < δ
=> 0< x <δ

|f(x) - L| < ϵ
| \(\sqrt{4-x}\) - 2 | < ϵ
- ϵ < \(\sqrt{4-x}\) - 2 < ϵ
2-ϵ < \(\sqrt{4-x}\) < ϵ + 2
(2-ϵ)^2 < 4-x < (ϵ + 2)^2
ϵ^2 - 4ϵ < -x < ϵ^2 + 4ϵ
4ϵ - ϵ^2 > x > -ϵ^2 - 4ϵ

Then, I'm lost!

Answer 2

Ummmmm I don't think you want to square each term like that, I think it'll get you into more trouble D: I think you wanna go about it like this somehow... lemme do a few more steps.. see if i remember how to do this correctly.