On Friday, July 11, 2014 12:35:45 AM UTC+2, dull...@sprynet.com wrote: > On Thu, 10 Jul 2014 10:19:11 -0700 (PDT), John Gabriel > > <email@example.com> wrote: > > > > >On Thursday, July 10, 2014 6:56:19 PM UTC+2, dull...@sprynet.com wrote: > > > > > >> >> >> No. Saying N=eps-|f(x)-L| is nonsense, because N is not allowed > > >> >> >> to depend on x. > > > > > >> >Nonsense. Allowing N=eps-|f(x)-L| proves that N works for all x>N. > > > > > >> N is not allowed to depend on x. That's simply how the logic of "Ae EN (Ax ...) works. > > > > > >There is nothing that says N cannot be stated in terms of x. > > > > > >The definition is: "for every eps > 0 there exists N such that |f(x) - L| < eps for all x > N". > > > > _That_ is exactly what says N cannot depend on x. First "there exists > > N such that", and then something about x. N is chosen _first_, > then given N it's suppoed to be true that something happens > for every x > N.
> Look. Is the following true or false? > (*) There exists N such that every x is larger than N.
> I hope you agree that (*) is false. No N is larger than > _every_ x.
> But by your (wrong) logic it's easy to show that > (*) is true: Just let N = x - 1.
Noooo, nooo, noooo. You are NOT understanding this problem correctly. Setting N=eps-|f(x)-L| helps to prove the result. x is NOT fixed, therefore, you are not constraining the choice of N. You seem to miss the fine detail here, which is not surprising!
> If N = x - 1 then x > N. True, Does that say > that there exists N which is larger than every x? > No. It seys for every x there exists N which > is larger than x. Very different statements.
> But if N is allowed to depend on x then the > clear false statment (*) becomes true. > N is simply not allowed to depend on x.
Yes, it is. There is NOTHING that prevents me from setting N=eps-|f(x)-L|.
> For the same reason as in the definition of limit.