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Limit, and lopitals rule

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Find A so that lim (x+A/x-2A)^x =5
    x-> 00

    2. Relevant equations



    3. The attempt at a solution
    gotta use lopihtals rule.
     
  2. jcsd
  3. May 16, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi gayani! Welcome to PF! :smile:

    (have an infinity: ∞ and try using the X2 tag just above the Reply box :wink:)
    hmm … I can't see how to use l'Hôpital's rule :redface:

    but if you write the LHS as (1 + 3A/(x - 2A))x, and fiddle about with it, you should get something you recognise. :smile:
     
  4. May 16, 2009 #3

    Mark44

    Staff: Mentor

    Gayani,
    Just in case you don't follow what Tiny-tim did, here it is.

    You wrote the rational expression as x + A/x - 2A. What I'm pretty sure you meant (and the way Tiny-tim interpreted it) was (x + A)/(x - 2A). What you wrote would be correctly interpreted as the sum and difference of three expressions-- x, A/x, and 2/A, not the quotient of x + A and x - 2A.

    What Tiny-tim did was divide x + A by x - 2A, so that (x + A)/(x - 2A) = 1 + 3A/(x - 2A).
     
  5. May 16, 2009 #4
    Take the natural log of both sides and then take the limit of the natual log of the left side

    [tex]\lim_{x\rightarrow\infty} ln\left(1 + \frac{3A} {x-2A}\right)^x= ln (5)[/tex]

    You'll have to do some rearranging in order to use l'Hôpital's rule and "solve" for A.
     
    Last edited: May 16, 2009
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