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Prove if abs|f(x)| <= B for all x#0, then lim as x->0 of xf(x) = 0,

  1. Oct 13, 2011 #1
    1. if there is a number B such that abs|f(x)| <= B for all x#0, then lim as x->0 of xf(x) = 0,


    2. Would I be able to get some help on how to prove this?


    3. Given Epsilon > 0 such that delta = epsilon/B will prove the limit equals 0. (I know that may not make much sense but how am I supposed to make a solution when I don't know where to start... but I was warned so here is my attempt.) :|
     
    Last edited: Oct 13, 2011
  2. jcsd
  3. Oct 13, 2011 #2

    SammyS

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    Welcome to PF .

    It looks like a fairly direct δ-ε proof.

    What have you tried?

    Where are you stuck?
     
  4. Oct 13, 2011 #3
    Thanks, the thing is I don't know where to start. I mean my teacher always starts out with given ε > 0 such that δ... I honestly just don't know where to go either.
     
  5. Oct 13, 2011 #4

    Mark44

    Staff: Mentor

    I think your teacher starts out with a given ε > 0, and then figures out what δ needs to be.

    Look at your text or class notes for some examples.
     
  6. Oct 13, 2011 #5
    Oh, yea, thanks! I didn't think of that! :surprised
     
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