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Lim (x-> infinity) sinh(x)sinh(e^(-x))

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data
    lim (x-> infinity) sinh(x)sinh(e^(-x))


    2. Relevant equations
    None really.


    3. The attempt at a solution
    L'Hospital?
     
  2. jcsd
  3. May 23, 2009 #2

    Dick

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    Re: Limit

    l'Hopital, yes. Try writing it as sinh(e^(-x))/(1/sinh(x)). Now the form is 0/0.
     
  4. May 23, 2009 #3
    Re: Limit

    hmm.. still failing at this question. pls help further?
    I don't get a nice solution on applying lhospital
     
  5. May 23, 2009 #4

    Dick

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    Re: Limit

    What did you get from l'Hopital?
     
  6. May 23, 2009 #5
    Re: Limit

    i now get the limit to infinity of:
    e^(-x)cosh(e^(-x))/(cothxcosechx)
    considering substitution of y=f(x) then computing the limit for lny
     
  7. May 23, 2009 #6

    Dick

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    Re: Limit

    That looks ok. Now look at the parts. What's lim coth(x)? What's lim cosh(e^(-x))?
     
  8. May 23, 2009 #7
    Re: Limit

    cothx -> 1
    cosechx -> 0
    cosh(e^-x) -> 1
    e^-x -> 0
    i guess lhopital again...
     
  9. May 23, 2009 #8

    Dick

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    Re: Limit

    Not so fast. Aside from the stuff that goes to 1, you've got e^(-x)*sinh(x). What's that? Use the definition of sinh.
     
  10. May 23, 2009 #9
    Re: Limit

    sweet. so it's 1/2.
     
  11. May 23, 2009 #10

    Dick

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    Re: Limit

    Not so bad, huh?
     
  12. May 23, 2009 #11
    Re: Limit

    Nope. Must be getting tired... it's like midnight here.
     
  13. May 23, 2009 #12
    Re: Limit

    Also could u please verify that limit (x,y) -> (0,0) of (x^4*y^2)/(x^2 + y^2)2 is 0
     
  14. May 23, 2009 #13

    Dick

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    Re: Limit

    Express the function in polar coordinates. Count powers of r.
     
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