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!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Absolutely! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook