Quote by PirateFan308
Another question, is there more than one δ that will prove this?
Say, Given an arbitrary ε>0, take δ=min{2,ε/6}
If x≠2 and x2<δ, then x2<2 and x2<ε/6
f(x)L = x^{2}4 = (x+2)(x2) = x+2x2
x2<2 => 2<x2<2 => 0<x<4 => 2<x+2<6 => x+2<6
x2x+2 < (6)(ε/6) = ε so f(x)L<ε

Sure, that choice works just as well.