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Help w/ limit

  1. Dec 13, 2004 #1
    lim as x approaches 1 of x raised to the quantity 1/1-x. When I first did it, I got 1 as my answer, but I have a strong feeling the answer is 0 also. I don't know how to get it, however. 1 raised to any power is 1, that is where I got the answer 1 from.
  2. jcsd
  3. Dec 13, 2004 #2
    try l'hopital's rule. take ln of your function, then find the limit, then exponentiate to undo the ln

    edit: the answer i got is 1/e
    Last edited: Dec 13, 2004
  4. Dec 13, 2004 #3


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    I got infinity as the answer :grumpy:.

    [tex]x^{\frac{1}{1-x}} = e^{\frac{lnx}{1-x}}[/tex]


    [tex]\lim_{x \rightarrow 1} \frac{lnx}{1-x} = \lim_{x \rightarrow 1} \frac{1}{x(1-x)^2} = \infty [/tex]
  5. Dec 13, 2004 #4


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    How did you get
    [tex]\lim_{x \rightarrow 1} \frac{lnx}{1-x} = \lim_{x \rightarrow 1} \frac{1}{x(1-x)^2} = \infty [/tex]
    Using L'Hopital, you would take the limit of [itex]\frac{1/x}{-1}[/itex] as x-> 1 which is -1. The limit of [itex]x^{\frac{1}{1-x}}[/itex] is e-1 as fourierjr said.
  6. Dec 13, 2004 #5
    That makes sense... I never even thought yo take the ln of x or use L. Hospital's rule. Thanks guys
  7. Dec 13, 2004 #6
    You dont have to use L'Hopital

    look closely to your limit

    [tex]\lim_{x \rightarrow 1} \frac{\log x}{1-x} = \lim_{x \rightarrow 1} \frac{\log x-\log 1}{1-x}[/tex]

    wich is the definition of the derivative of [itex]-\log x[/itex] evaluated in [itex]1[/itex] so its obvious than the limit is [itex]-1[/itex].

    Much more elegant dont you think?
    Last edited: Dec 13, 2004
  8. Dec 14, 2004 #7


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    Not than my version :approve:
    [tex] lim_{x\rightarrow 1} x^{\frac{1}{1-x}} =lim_{y\rightarrow 0} (1-y)^{\frac{1}{y}}=\frac{1}{e} [/tex]
    ,where i made use of a simple substituion... :tongue2: and of a very known limit. :wink:

  9. Dec 14, 2004 #8
    nice and simple !!
  10. Dec 14, 2004 #9
    hey Daniel, can you remind me of how

    [tex]\lim_{y\rightarrow 0} (1-y)^{\frac{1}{y}}[/tex]

    is calculated???? too lazy to open my book :P
  11. Dec 14, 2004 #10


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    Kay,i'll make an exception:
    [tex]\lim_{y\rightarrow 0}(1-y)^{\frac{1}{y}}=\lim_{x\rightarrow +\infty}(1-\frac{1}{x})^{x} =\frac{1}{e} [/tex]

  12. Dec 14, 2004 #11
    Nice solution, dextercioby!
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