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Homework Help: Prove lim x/(x+1) = 1/2

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove
    lim x/(x+1) = 1/2
    x->1

    2. Relevant equations

    3. The attempt at a solution
    |x/(x+1) - 1/2|=|x-1|/(2|x+1|)
    Assume x>0 (can I say this??), then |x-1|/(2|x+1|)<|x-1|/2
    Take delta=min{1,epsilon/2}.
    Then if 0<|x-1|<delta, then |x/(x+1) - 1/2|<epsilon

    In the middle of my proof, I assumed that x>0, is this OK?
    Does my choice of delta (delta=min{1,epsilon/2}) work?

    Could someone kindly confirm this? (or point out any mistakes)
    I haven't done those for awhile, so I'm not sure if I'm doing it correctly.
    Thanks!
     
  2. jcsd
  3. Feb 26, 2010 #2

    tiny-tim

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    Hi kingwinner! :smile:

    (have a delta: δ and an epsilon: ε :wink:)

    I don't like your x > 0 … it's messy, and you really ought to do x < 0 also. :redface:

    You're trying to get a lower limit on |x+1|, and you've chosen 2 … which doesn't really work for x < 0, does it?

    try something less than 2, eg 1.5 :wink:
     
  4. Feb 26, 2010 #3
    hmm...so my "proof" above is wrong? :(

    But I thought the definition of limit says as x gets close to 1 (0<|x-1|<delta), then the function gets arbitrarily close to the limit 1/2. We don't really have to care at all about the case x<0, right? If I take delta=min{1,epsilon/2}, then delta is always less than or equal to 1, so we never have to look at the case x<0, isn't it?

    So if I say I assume x>0 in the first place of the proof. Is this ok, can somebody confirm or correct me, please?
     
    Last edited: Feb 26, 2010
  5. Feb 26, 2010 #4

    arildno

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    tiny-tim!

    I totally disagree with you!

    Being very close to x=1 implies x>0, so there is no need to consider x<0.

    kingwinner:

    Under your assumption x>0 (perfectly okay!), you have derived the bound <|x-1|/2

    Thus, by chosing delta = 2e, whenever |x-1|<delta, your expression will be less than e.
     
  6. Feb 26, 2010 #5
    x>0 => 1- delta > 0 => delta <1 so I think here is the requirement in addition to the bound I obtained later on.

    So I think if I take delta=min{1, 2e}, then both conditions hold, and if 0<|x-1|<delta, then |x/(x+1) - 1/2|<epsilon ? I don't think it's enough to just take delta=2e...we need delta to be no greater than 1 as well, right?
     
  7. Feb 26, 2010 #6

    tiny-tim

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    Hi arildno!

    Yes, you're right, I was confused about what x was (also I missed the factor of 2 in 1/2|x+1|). :redface:

    Thanks for the correction. :smile:
     
  8. Feb 26, 2010 #7
    I actually originally meant to take delta=min{1, 2e}, but I made a silly computational mistake and got e/2 instead of 2e...

    But if I really take delta=min{1, e/2}, it's going to work as well because if a certain delta works in the definition, than any smaller delta (>0) will certainly work, right?
     
  9. Feb 26, 2010 #8

    arildno

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    Absolutely! :smile:
     
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