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Nov13-11, 07:27 PM
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Quote Quote by PirateFan308 View Post
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 |x-2|<δ, then |x-2|<2 and |x-2|<ε/6
|f(x)-L| = |x2-4| = |(x+2)(x-2)| = |x+2||x-2|
|x-2|<2 => -2<x-2<2 => 0<x<4 => 2<x+2<6 => |x+2|<6
|x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε
Sure, that choice works just as well.