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Thread: Limit

  1. #1
    Senior Member pankaj's Avatar
    Jul 2008
    New Delhi(India)


     <br />
\lim_{x\to 0+}(x^{x^x}-x^x)<br />
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  2. #2
    MHF Contributor Mathstud28's Avatar
    Mar 2008
    Quote Originally Posted by pankaj View Post
     <br />
\lim_{x\to 0+}(x^{x^x}-x^x)<br />
    There are eight million ways to do these, Ill show a couple. First let us handle \lim_{x\to{0}}x^x

    Way One: if we let x=\frac{1}{\varphi} this limit transforms to \lim_{\varphi\to\infty}\left(\frac{1}{\varphi}\rig  ht)^{\frac{1}{\varphi}}, now by the connection of the Root and Ratio tests, this is equivalent to \lim_{\varphi\to\infty}\frac{\frac{1}{\varphi+1}}{  \frac{1}{\varphi}}=1

    Way Two: After the sub \frac{1}{\varphi}\left(\frac{1}{\varphi}\right)^{\  frac{1}{\varphi}} we may rewrite it as \lim_{\varphi\to\infty}e^{\ln\left(\left(\frac{1}{  \varphi}\right)^{\frac{1}{\varphi}}\right)}=\lim_{  \varphi\to\infty}e^{\frac{-\ln(\varphi)}{\varphi}} and by the continuity of the exponentiat function across its entire domain we may rewrite this as e^{\lim_{\varphi\to\infty}\frac{-\ln(\varphi)}{\varphi}}. Now by any number of methods (I leave this to you) \lim_{\varphi\to\infty}\frac{\ln(\varphi)}{\varphi  }=0 so we have our limit is equal to e^0=1

    Now my fingers are getting tired so lets only consider one method for this one. Once again let us rewrite our limit as \lim_{x\to{0^+}}e^{\ln\left(x^{x^x}\right)}=e^{\li  m_{x\to{0}}x^x\cdot\ln(x)}

    Now x^x\to{1} and \ln(x)\to-\infty

    So we have that x^x\cdot\ln(x)\to-\infty and that makes e^{\lim_{x\to{0^+}}x^x\ln(x)}\to{0}

    So the asnwer to your actual limit is 0-1=-1
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