Originally posted by humy
oh I see. Obviously cannot have |f(z)| < 0 .
I will revise my formula and then came back here later.
So, if I am at LAST thinking about this correctly, the problem with that formula:
lim {x→∞} f(x) = L ⇒ ∀x ∈ ℝ ∃z ∈ ℝ : z>x ∧ |f(z) − L| < |f(x) − L| ( FOUL! )
( " {...} " indicates subscript )
is that, if f(x) is an oscillating function, as x increases, f(x) may 'pass through' f(x) = L not once but many times and thus, if we make the value of x such that f(x) = L where x is such that that L can be reached yet again by increasing x by a sufficient non-zero value yet again, then we cannot have;
|f(z) − L| < |f(x) − L| as the closest we can get to that is |f(z) − L| = |f(x) − L| .
Now, if I am at last thinking about this correctly, a fix for that would be to simply replace that "<" with "≤" so that we can allow that closest approach of |f(z) − L| = |f(x) − L| and that means changing the formula to;
lim {x→∞} f(x) = L ⇒ ∀x ∈ ℝ ∃z ∈ ℝ : z>x ∧ |f(z) − L| ≤ |f(x) − L|
For similar reason, I got it a bit wrong for;
lim {x→p} f(x) = L ⇒ ∀x ∈ ℝ{≠p} ∃z ∈ ℝ : |z − p| < |x − p| ∧ |f(z) − L| < |f(x) − L| ( FOUL! )
and that should be;
lim {x→p} f(x) = L ⇒ ∀x ∈ ℝ{≠p} ∃z ∈ ℝ : |z − p| < |x − p| ∧ |f(z) − L| ≤ |f(x) − L|
Anyone:
Have I at last got both of those above formulas exactly correct?