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L'hospitals rulee major help

  1. Mar 14, 2007 #1
    l'hospitals rulee!!! major help

    1. i don't know how to do this question
    limx->infinity (x/x+1)^x. i keep gettin the rong answer, some heelp plzz:confused:




    2.the right answer is 1/e but im not getting that, i tried many diff ways, i cant post my work it here because its too long and confusing



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 14, 2007 #2
    Hint

    Consider the series expansion for e^[1/x], when x is large.
     
  4. Mar 14, 2007 #3
    umm so when x is large 1/x = 0 so that means e^o = 1, im still lossttt
     
  5. Mar 14, 2007 #4
    Do you know the Taylor series for e^y, when y is small?
     
  6. Mar 14, 2007 #5
    noo..havent learned it =S
     
  7. Mar 14, 2007 #6
    i took the ln of both sides, and then found the derivative, and i got the answer as zero.. so i got ln y= 0
    den i used the fact that e^lny = y...there fore y= e^0 =S..
     
  8. Mar 15, 2007 #7
    Hmmm, I'm not sure how to do it without recourse to a Taylor series expansion. It's easy to prove though that

    e^x=1+x+x^2/2!+x^3/3+....

    Have you done Taylor series at all?
     
  9. Mar 15, 2007 #8
    noo i haven't :|
     
  10. Mar 15, 2007 #9
    kk thnnxx for your help newayzz =)
     
  11. Mar 15, 2007 #10
    You suggested that L'hopital's rule should be used in the derivation. I played around with that and couldn't see it, but maybe someone else can work it out.

    Let me know when you get the answer.
     
  12. Mar 15, 2007 #11

    Dick

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    You should have gotten something like x*ln(x/(x+1)) when you took the log. This is an infinity*0 form. To use L'Hopital you want to write it as 0/0 (or infinity/infinity). So write it as ln(x/(x+1))/(1/x). Now take derivative of the numerator and denominator and send x->infinity.
     
  13. Mar 15, 2007 #12

    dextercioby

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    There's no need for l'Hospital's rule, as this limit is elementary

    [tex] \lim_{x\rightarrow \infty}\left(1-\frac{1}{x+1}\right)^{(x+1)\frac{x}{x+1}} =e^{-\lim_{x\rightarrow \infty}\frac{x}{x+1}}=e^{-1} [/tex]
     
  14. Mar 15, 2007 #13
    Dick, Brilliant!
     
  15. Mar 15, 2007 #14
    That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.
     
  16. Mar 15, 2007 #15

    Dick

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    Thanks. Wish I could say I invented it, but it's a pretty routine thing to do.
     
  17. Mar 15, 2007 #16

    dextercioby

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    Pardon me ? I just the definition of "e". What's more elementary than that ??:rolleyes:
     
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