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Proof that a limit is less than or equal to zero

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Proof that, If [itex]f[/itex] is a function such that

    (1) [itex]f[/itex] is differentiable at (open) the interval [itex]D[/itex],

    (2) [itex]D[/itex] includes [itex]0[/itex] and [itex]f(0)=0[/itex], and

    (3) for all [itex]x[/itex] in [itex]D[/itex] other than [itex]0[/itex], [itex]f(x)[/itex] and [itex]x[/itex] have opposite signs

    Then

    [itex]f'(0)\leq0[/itex]

    2. Relevant equations

    None.

    3. The attempt at a solution

    I managed to prove that for all [itex]x[/itex] in [itex]D[/itex] other than [itex]0[/itex]

    [itex]\frac{f(x)-f(0)}{x-0}\leq0[/itex]

    I don't know how to get from there to the fact that

    [itex]lim _{x\rightarrow0} \frac{f(x)-f(0)}{x-0}\leq0[/itex]


    Any help would be very appreciated. Thanks.
     
    Last edited: Sep 1, 2013
  2. jcsd
  3. Sep 1, 2013 #2
    If for some function [itex]g(x)[/itex] we have [itex]lim _{x\rightarrow0}g(x)=L>0[/itex], then can you argue that [itex]g(x)[/itex] must be positive in some (-δ,δ)\{0} by considering a certain ε>0?
     
    Last edited: Sep 1, 2013
  4. Sep 1, 2013 #3
    Okay, I got it. If the limit equals [itex]k[/itex] and [itex]k>0[/itex], then

    [itex]\forallε>0\:\existsδ>0 (|x|<δ \rightarrow \left|\frac{f(x)-f(0)}{x-0}-k\right|<ε)[/itex]

    implies (for [itex]ε=k/2[/itex]) that

    [itex]\existsδ>0 (|x|<δ \rightarrow \frac{k}{2}<\frac{f(x)-f(0)}{x-0}<\frac{3k}{2})[/itex]

    But [itex]\frac{k}{2}<\frac{f(x)-f(0)}{x-0}<\frac{3k}{2}[/itex] cannot be true at [itex](0,δ)[/itex], because it would contradict statement (3).

    Thanks!
     
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