1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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



    2. Relevant equations


    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]


    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).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Proof that a limit is less than or equal to zero
  1. Proof about less than (Replies: 1)