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Math Help - lim x^x

  1. #1
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    lim x^x

    with proof find the lim_{x -> 0+}(x^{x})

    my guess would be 1 as it is between 0^0< x^x < 1^x for  0 < x < 1 and by the sandwich rule it convergrs to 1
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  2. #2
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    Hi

    Quote Originally Posted by rosh3000 View Post
    with proof find the lim_{x -> 0+}(x^{x})
    I would solve it this way:

    lim \ x^x = lim \ e^{ln(x)*x} = e^0 = 1


    Quote Originally Posted by rosh3000 View Post
    my guess would be 1 as it is between 0^0< x^x < 1^x for  0 < x < 1 and by the sandwich rule it convergrs to 1
    I have never heard about this rule before. Do you have to use it?

    Yours
    Rapha
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  3. #3
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    The sandwich theorem says :

    a_n\leq b_n\leq c_n and \lim_{n\to\infty} a_n=\lim_{n\to\infty} c_n=l \quad \Rightarrow \lim_{n\to\infty} b_n=l

    What you used is false and not the sandwich theorem...
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  4. #4
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    Quote Originally Posted by rosh3000 View Post
    with proof find the lim_{x -> 0+}(x^{x})

    my guess would be 1 as it is between 0^0< x^x < 1^x for 0 < x < 1 and by the sandwich rule it convergrs to 1
    The limit is of the indeterminate form 0^0

    So we will 1st take the log of x^x and then apply L'Hospital's rule.

    AND

    \lim_{x\rightarrow 0^+}{\ln x^x} = \lim_{x\rightarrow 0^+}({x\ln x}) = \lim_{x\rightarrow 0^+}{\frac{\ln x}{\frac{1}{x}}} = \lim_{x\rightarrow 0^+}{\frac{\frac{1}{x}}{\frac{-1}{x^2}}} = \lim_{x\rightarrow 0^+}({-x}) = 0.

    Since \ln x^x tends to 0, x^x must tend to 1.
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  5. #5
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    thanks you mak it seem so easy
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  6. #6
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    Quote Originally Posted by Rapha View Post
    Hi



    I would solve it this way:

    lim \ x^x = lim \ e^{ln(x)*x} = e^0 = 1






    Yours
    Rapha

    Does not the product (ln x)*x ,as x goes to zero go to -\infty*0,which is indeterminent,if im not mistaken??
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  7. #7
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    \lim_{x\to 0}x\ln(x)=0. You can either do this by looking at the Taylor series for \ln(x) or by rewriting it as \frac{\ln(x)}{\frac{1}{x}} and using L'Hopital's.
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  8. #8
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    Quote Originally Posted by xalk View Post
    Does not the product (ln x)*x ,as x goes to zero go to -\infty*0,which is indeterminent,if im not mistaken??
    Huh? That is weird, because you already posted the solution yourself

    Quote Originally Posted by xalk
    \lim_{x\rightarrow 0^+}({x\ln x}) = \lim_{x\rightarrow 0^+}{\frac{\ln x}{\frac{1}{x}}} = \lim_{x\rightarrow 0^+}{\frac{\frac{1}{x}}{\frac{-1}{x^2}}} = \lim_{x\rightarrow 0^+}({-x}) = 0


    This is exactly what you asked for, and you gave the answer in your post before. Quite funny
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  9. #9
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    Hey, just another slightly different approach without using L'Hopital:

    By considering their graphs it's easy to see that

    <br /> <br />
\lim_{a \to - \infty} \int_{a}^0 \ e^x dx = - \lim_{b \to 0} \int_{b}^1 \ \ln{x} dx <br />

    Then the left hand side is easy to calculate:
    <br />
\lim_{a \to - \infty} \int_{a}^0 \ e^x dx = \lim_{a \to - \infty} [e^x]_{a}^0 = 1<br />
    So on the right hand side

    <br />
\lim_{b \to 0} \int_{b}^1 \ \ln{x} dx = -1

    <br />
\lim_{b \to 0} [x \ln{x} - x]_{b}^1 = -1<br />

    rearraging we arrive at

    <br />
\lim_{x \to 0} x \ln{x} = 0<br />
\Rightarrow \lim_{x \to 0} x^x = 1<br />

    Using L'Hopital is clearly a lot quicker

    Hope this was useful.
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  10. #10
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    Another way to view it, is remembering that if x \rightarrow 0 \ln(x) \rightarrow x-1.

    So \lim_{x \to 0} x^x=\lim_{x \to 0} e^{x \ln (x)}=\lim_{x \to 0} e^{x (x-1)} = 1
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  11. #11
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    Quote Originally Posted by Ruun View Post
    Another way to view it, is remembering that if x \rightarrow 0 \ln(x) \rightarrow x-1.
    False.
    You would get that \lim_{x\to 0} \ln(x)=-1
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  12. #12
    Member Ruun's Avatar
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    You're right, didn't thought twice. Thanks for the correction!
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  13. #13
    MHF Contributor chisigma's Avatar
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    Some years ago I demonstrated in rigorous way that the expression 0^{0} is not at all 'undetermined' but it is...

    0^{0}=1

    That meas that ...

    f(x)= x^{x}

    ... is continous in x=0 because the limit and the value of the function are identical...

    Kind regards

    \chi \sigma
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  14. #14
    Moo
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    Quote Originally Posted by chisigma View Post
    Some years ago I demonstrated in rigorous way that the expression 0^{0} is not at all 'undetermined' but it is...

    0^{0}=1
    I like the way you refer to yourself... It would be more interesting to give an outline of the proof... (I've seen some, I'm thinking about anyone who reads this)

    That meas that ...

    f(x)= x^{x}

    ... is continous in x=0 because the limit and the value of the function are identical...
    And the limit as x comes from the left : x\to 0^- ?
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  15. #15
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Moo View Post
    I like the way you refer to yourself... It would be more interesting to give an outline of the proof... (I've seen some, I'm thinking about anyone who reads this)
    As Dante Alighieri wrote '... perder tempo a chi pił sa pił spiace...', so that the 'proof' I'll give to you is quite elementary...

    phase 1: you activate the function 'calculator' in the 'accessories' of Windows...

    phase 2: you set the calculator as 'scientific' ...

    phase 3 : push in order '0' -> 'x^y' -> '0' -> '='...

    After phase 3 in the display of calculator you will read 1, so that...

    0^{0}=1

    Please avoid tedious further discussions ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; June 15th 2009 at 06:51 AM. Reason: spelling
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