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Problem with Limits using L'Hospital's Rule

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the limit of

    lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 of [tex]\psi[/tex]^(1/([tex]\psi[/tex]-1))


    2. Relevant equations
    psigraph.jpg


    3. The attempt at a solution

    Let y = [tex]\psi[/tex]^(1/([tex]\psi[/tex]-1))
    ln y = ln [tex]\psi[/tex]^(1/([tex]\psi[/tex]-1))

    lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 ln y = lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 of (1/([tex]\psi[/tex]-1)) (ln [tex]\psi[/tex])

    Differentiate

    lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 ln y = -1/([tex]\psi[/tex]-1)2 x (1/[tex]\psi[/tex])

    lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 ln y = 2/([tex]\psi[/tex]3+3[tex]\psi[/tex]2+3[tex]\psi[/tex]+1)

    ln y =1/4
    y = e1/4

    Does e1/4 = e?
     
  2. jcsd
  3. Feb 28, 2009 #2
    L'Hopital's rule works for f(x)/g(x) and then you get f'(x)/g'(x). Try rewriting the step before you differentiate as a fraction and not a product.
     
  4. Feb 28, 2009 #3
    Thank you for your reply.

    I wonder which is the correct solution

    solution (a)

    ln y = 1/([tex]\psi[/tex]-1) x ln [tex]\psi[/tex]

    ln y = ln [tex]\psi[/tex] x ([tex]\psi[/tex]-1)

    ln y = 1/[tex]\psi[/tex]

    ln y = 1/1

    y = e1

    or solution (b)

    ln y = 1/([tex]\psi[/tex]-1) x ln [tex]\psi[/tex]

    ln y = ln [tex]\psi[/tex] / ([tex]\psi[/tex]-1)

    ln y = 1/[tex]\psi[/tex]

    ln y = 1/1

    y = e1
     
    Last edited: Feb 28, 2009
  5. Feb 28, 2009 #4
    I did it the second way, assuming you just didn't feel like typing out that you were still dealing with limits
     
  6. Feb 28, 2009 #5
    Thank you Hogger for your point outs. Appreciate it. Have a nice day



    Charles
     
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