1. Not finding help here? Sign up for a free 30min 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!

Limits of modulus

  1. Jun 1, 2014 #1
    Proof that [tex] \lim_{x→p} \frac {f(x)}{x - p} = 0 \Leftrightarrow \lim_{x \rightarrow p} \frac{f(x)}{\left | x - p\right |} = 0 [/tex]

    Relevant equations
    I think there is a useful theorem here, the conservation of signal:
    Suppose that [itex] \lim_{x\rightarrow p} f(x) = L [/itex]; [itex] L > 0 [/itex]; [itex]\forall \varepsilon >0 [/itex], there is [itex]\delta > 0[/itex] so that [itex]\forall x \in D_f[/itex], [tex] 0 < \left | x - p \right | < \delta \Rightarrow f(x) > 0[/tex].



    The attempt at a solution
    I've figured out a solution for the direct implication, and it's very easy:
    [tex] \lim_{x\rightarrow p} \frac{f(x)}{x - p} = 0 \Rightarrow \forall \varepsilon >0 \exists \delta > 0 : 0 < \left | x - p \right | < \delta \Rightarrow \left | \frac{f(x)}{x - p} \right | < \varepsilon.[/tex]
    As [itex]\left | x - p \right | = \left | \left | x - p \right | \right |[/itex], the proof is over (at least, this is my opinion). The problem arise when I try to proof the reverse implication:
    [tex] \lim_{x \rightarrow p} \frac{f(x)}{\left | x - p\right |} = 0 \Rightarrow \lim_{x→p} \frac {f(x)}{x - p} = 0 . [/tex]
    That's because I never considered elegant to use "reverse engeneering" to prove reverse implications (I mean, use the same strategy for both). For me, this is not elegant, and we must use another strategy to prove the reverse of the theorem, but I am with no ideas for this. Anyone has a clue of an independent method to prove it?

    Thanks in advance!
     
    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 1, 2014 #2
    Well, maybe a simple replacement of the symbol [itex] \Rightarrow [/itex] for [itex] \Leftrightarrow [/itex] in the proof above should solve the problem. I've seen it in some caculus textbooks. What do you think, guys?
     
    Last edited: Jun 1, 2014
  4. Jun 1, 2014 #3

    pasmith

    User Avatar
    Homework Helper

    Hint: a two-sided limit exists if and only if both one-sided limits exist and are equal.
     
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: Limits of modulus
  1. Finding the modulus (Replies: 2)

  2. Modulus of z (Replies: 11)

  3. Modulus Question (Replies: 4)

Loading...