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L'hospital's Rule

  1. Nov 28, 2007 #1
    1. The problem statement, all variables and given/known data

    lim[tex]_{x -> infin}[/tex] x[(x+1)ln(1+(1/x))-1]



    3. The attempt at a solution

    Am I able to start this with the chain rule. I'm not sure how to start this.
     
  2. jcsd
  3. Nov 28, 2007 #2
    i've been simplifying your original problem to see what form of L'Hospital's you have

    [tex]\lim_{x \rightarrow \infty} x[(x+1)\ln(1+\frac{1}{x})-1][/tex]

    also is this correct?
     
  4. Nov 28, 2007 #3
    well that's the oringal problem
    Is it an infinity times infinity problem, but to use l'hospital's rule it has to be a fraction
     
  5. Nov 28, 2007 #4
    can i leave x on top and move [(x+1)ln(1+(1/x))-1] to the bottom by raising it to the -1 power? Then use l'hospital's rule
     
  6. Nov 28, 2007 #5

    nrqed

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    I don't know if this is going to be useful to you because I don;t know if you have seen that trick but I can get the answer quickly by using a Taylor expansion of [tex] ln(1+\epsilon) \approx 1 + \epsilon [/tex] for small epsilon. Then simple algebra leads dircetly to the answer.
     
  7. Nov 28, 2007 #6
    ok think i may have a solution

    lim x-> infin [tex]\frac{x}{[(x+1)ln(1+(1/x))-1]^-1}[/tex]

    then use l'hospital's rule and get

    [tex]\frac{1}{-[(x+1)ln(1+(1/x))-1]^-1 * [ln(1+(1/x)+(x+1)(1/1+(1/x))}[/tex]
     
  8. Nov 28, 2007 #7
    yea, i'm not reallu sure about that. I'm not sure if it will be acceptable
     
  9. Nov 28, 2007 #8

    Dick

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    Be careful here, you may be quickly lead to the wrong answer. You need to keep the quadratic term in the expansion as well. I got burnt by that once.
     
  10. Nov 28, 2007 #9
    you mean in my use of the chain rule?..is this the right approach though?
     
  11. Nov 28, 2007 #10

    Dick

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    This limit isn't even the same as the one you started out with. Write it as
    [(x+1)*ln(1+1/x)-1]/(1/x). Now it's 0/0 (though that's not completely clear until you show (x+1)*ln(1+1/x) -> 1. You can use l'Hopital but you've got to be careful.
     
  12. Nov 28, 2007 #11
    so after you use l'hospital's rule the first time you get

    [tex]\frac{ln(1+\frac{1}{x})+(x+1)(\frac{1}{1+\frac{1}{x}})}{\frac{-1}{x^{2}}}[/tex]

    and that's infin/0 ...right?
     
  13. Nov 28, 2007 #12

    Dick

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    You were the one who brought up the chain rule. The derivative of ln(1+1/x) is [1/(1+1/x)]*(-1/x^2). It's 0/0. Do some algebra on that before you differentiate again.
     
  14. Nov 28, 2007 #13
    wait..i the correct answer -1 since it equals infin/neg infin
     
  15. Nov 28, 2007 #14

    Dick

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    infin/neg infin is NOT -1. It's indeterminant. You need to differentiate again. In the end you will find a perfectly finite limit.
     
  16. Nov 28, 2007 #15
    how many time do you have to use l'hospital's rule..it's getting really messy
     
  17. Nov 28, 2007 #16
    after the second time i use it i have 1/0
     
  18. Nov 28, 2007 #17

    Dick

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    This is the last time. But clean up the algebra in the numerator before you differentiate again. Better to do it before than after.
     
  19. Nov 28, 2007 #18
    but it's 1/0 can I still use l'hospitals'?
     
  20. Nov 28, 2007 #19

    Dick

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    I get [ln(1+1/x)-1/x]/(-1/x^2) after I do the algebra from the first differentiation. That's 0/0, not 1/0.
     
  21. Nov 28, 2007 #20
    your numerator is wrong, her's the oringal again

    lim x-> infin x[(x+1)ln(1+(1/x))-1]
     
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