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  1. #1
    sss
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    limits

    how to find lim (-ln(x))^x
    x->0+
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sss View Post
    how to find lim (-ln(x))^x
    x->0+
    This can be done any number of ways. Here is one, first note that \exists{N}\backepsilon\forall{x\geqslant{N}}~\ln(x  )\leqslant{x^{\varphi>0}}. Here I will use \forall{x}>{0}~~\ln(x)\leqslant\sqrt{x}. If you would like to prove this to yourself try finding the absolute maximum of f(x)=\sqrt{x}-\ln(x) and noting its sign.

    So now here is what we have

    \begin{aligned}\lim_{x\to{0}}\ln\left(\frac{1}{x}\  right)^x&=\lim_{x\to{0}}e^{x\ln\left(\ln\left(\fra  c{1}{x}\right)\right)}\\<br />
&=\lim_{\varphi\to\infty}e^{\frac{\ln(\ln(\varphi)  )}{\varphi}}{\color{red}\star}\\<br />
&=\exp\left(\lim_{\varphi\to\infty}\frac{\ln(\ln(\  varphi))}{\varphi}\right){\color{red}\star\star}\e  nd{aligned}

    \color{red}\star was made by the substitution \frac{1}{x}=\varphi and \color{red}\star\star was done using the continuity of the exponential function.

    So now let us note something

    \forall{\varphi}>1\quad0\leqslant\frac{\ln(\ln(\va  rphi))}{\varphi}\leqslant\frac{\ln(\varphi)}{\varp  hi}\leqslant\frac{\sqrt{\varphi}}{\varphi}=\frac{1  }{\sqrt{\varphi}}

    Cutting out the uneccessary step gives us

    \forall\varphi>1\quad0\leqslant\frac{\ln(\ln(\varp  hi))}{\varphi}\leqslant\frac{1}{\sqrt{\varphi}}

    So this implies that

    \lim_{\varphi\to\infty}0\leqslant\lim_{\varphi\to\  infty}\frac{\ln(\ln(\varphi))}{\varphi}\leqslant\l  im_{\varphi\to\infty}\frac{1}{\sqrt{\varphi}}

    Now \lim_{\varphi\to\infty}0=\lim_{\varphi\to\infty}\f  rac{1}{\sqrt{\varphi}}=0

    So we can see that

    0\leqslant\lim_{\varphi\to\infty}\frac{\ln(\ln(\va  rphi))}{\varphi}\leqslant0

    Which by the Squeeze Theorem indicates that \lim_{\varphi\to\infty}\frac{\ln(\ln(\varphi))}{\v  arphi}=0


    So then \exp\left(\lim_{\varphi\to\infty}\frac{\ln(\ln(\va  rphi))}{\varphi}\right)=e^0=1

    So we can finally conclude that \lim_{x\to{0^+}}\ln\left(\frac{1}{x}\right)^x=1
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