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Math Help - Limit of Indeterminate form 1^(infinity)

  1. #1
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    Limit of Indeterminate form 1^(infinity)

    A question on my homework is: lim(x->1+) (x^(1/(1-x)) The given answer is 0.367879 and by inspection of a graphing calculator given graph I can see that this is indeed correct, however when I attempt to find the limit: I set it up as a logarithm: ln y = ln(x^(1/(1-x)) = (1/(1-x)) ln(x) = ln(x) / (1-x) Then I use L'Hopital's Rule (since it is now of the indeterminate form 0/0): (1/x) / 1 which yields 1. We must remember to undo turning it into a logarithm so: e^1 = e but e is not the answer. >:/ How am I screwing this up?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by InfinitePartsInHarmony View Post
    A question on my homework is: lim(x->1+) (x^(1/(1-x)) The given answer is 0.367879 and by inspection of a graphing calculator given graph I can see that this is indeed correct, however when I attempt to find the limit: I set it up as a logarithm: ln y = ln(x^(1/(1-x)) = (1/(1-x)) ln(x) = ln(x) / (1-x) Then I use L'Hopital's Rule (since it is now of the indeterminate form 0/0): (1/x) / 1 which yields 1. We must remember to undo turning it into a logarithm so: e^1 = e but e is not the answer. >:/ How am I screwing this up?
    Here is your mistake : (1-x)'={\color{red}-}1. Using this, you'll find that the answer is \mathrm{e}^{-1}=\frac{1}{\mathrm{e}}.
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  3. #3
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    Quote Originally Posted by InfinitePartsInHarmony View Post
    A question on my homework is: lim(x->1+) (x^(1/(1-x)) The given answer is 0.367879 and by inspection of a graphing calculator given graph I can see that this is indeed correct, however when I attempt to find the limit: I set it up as a logarithm: ln y = ln(x^(1/(1-x)) = (1/(1-x)) ln(x) = ln(x) / (1-x) Then I use L'Hopital's Rule (since it is now of the indeterminate form 0/0): (1/x) / 1 which yields 1. We must remember to undo turning it into a logarithm so: e^1 = e but e is not the answer. >:/ How am I screwing this up?
    You have made a slight typo:

    \frac{\ln(x)}{1-x}~\underbrace{\implies}_{L'Hopital's\ rule}~{\color{red} - }\frac{\frac1x}{1} which yields -1 if x approaches 1.

    Therefore the limit is e^{-1} = \frac1e
    Last edited by earboth; July 14th 2008 at 12:59 AM.
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