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Math Help - Indeterminate powers

  1. #1
    DBA
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    Indeterminate powers

    Find the limit:

    lim x -> 0+ x^x

    write as an exponential:

    x^x = e^(xlnx)

    I know: lim x -> 0+ xlnx = 0 (we did that earlier with the l'Hospital's Rule)

    so,
    lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
    = e^0
    = 1

    2 things I do not understand.
    First, how do I know that this is indeterminate?

    Second,
    lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
    = e^0
    Why do I take the limit of the exponent of e if the limit of the entire term is needed?
    Thus, is lim x -> 0+ e^(xlnx) the same as e^lim x -> 0+ (xlnx) ??

    Thanks!
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  2. #2
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    Quote Originally Posted by DBA View Post
    Find the limit:

    lim x -> 0+ x^x

    write as an exponential:

    x^x = e^(xlnx)

    I know: lim x -> 0+ xlnx = 0 (we did that earlier with the l'Hospital's Rule)

    so,
    lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
    = e^0
    = 1

    2 things I do not understand.
    First, how do I know that this is indeterminate?


    What or who is indeterminate? You said you already evaluated the limit \lim_{x\to\infty}x\ln x by L'Hospital, and you did so because this is, of course, and indeterminate...


    Second,
    lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
    = e^0
    Why do I take the limit of the exponent of e if the limit of the entire term is needed?
    Thus, is lim x -> 0+ e^(xlnx) the same as e^lim x -> 0+ (xlnx) ??


    Indeed, and you can do this because the exponential function is continuous EVERYWHERE and thus evaluating a limit is, simply, substituing...

    Tonio


    Thanks!
    .
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  3. #3
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    Solve Like This :

    We Know that :


    <br /> <br />
 x =e ^ {\ln x}<br /> <br />

    So

    <br />
\lim_{x\rightarrow 0}e^{{\ln x}^x}= \lim_{x\rightarrow 0}e^{x\ln x}=e ^{\lim_{x\rightarrow 0}x{\ln x}}=e ^{\lim_{x\rightarrow 0}\frac {\ln x}{\frac {1}{x}}}\rightarrow {Lopital} = e^0 = 1<br />
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  4. #4
    DBA
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    Quote Originally Posted by tonio View Post
    .
    What I did not understand is how do I know that

    lim x -> 0+ x^x is indeterminate.

    Normally I would check the limit of the term and the exponent...

    lim x -> 0+ x = ? and lim x -> 0+ x = ? Then I would get a type like 0/0 or infinite/infinite. If I have a type like that than I am allowed to use the l'Hospital's rule.

    In this case I wasn't sure if I can use this check too? So, my question is: What indeterminate type do I have for lim x -> 0+ x^x
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  5. #5
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    Quote Originally Posted by DBA View Post
    What I did not understand is how do I know that

    lim x -> 0+ x^x is indeterminate.

    Normally I would check the limit of the term and the exponent...

    lim x -> 0+ x = ? and lim x -> 0+ x = ? Then I would get a type like 0/0 or infinite/infinite. If I have a type like that than I am allowed to use the l'Hospital's rule.

    In this case I wasn't sure if I can use this check too? So, my question is: What indeterminate type do I have for lim x -> 0+ x^x

    x^x=e^{x\ln x} and x\ln x = \frac{\ln x}{1\slash x} is the indeterminate form when  x\rightarrow 0 ...
    After solving this indeterminate you just substitute in the exponential because, as already pointed out, the exp function is a continuous one.

    Tonio
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  6. #6
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    Quote Originally Posted by DBA View Post
    Find the limit:

    lim x -> 0+ x^x

    write as an exponential:

    x^x = e^(xlnx)

    I know: lim x -> 0+ xlnx = 0 (we did that earlier with the l'Hospital's Rule)

    so,
    lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
    = e^0
    = 1

    2 things I do not understand.
    First, how do I know that this is indeterminate?
    http://www.math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf
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