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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 $\displaystyle \exists{N}\backepsilon\forall{x\geqslant{N}}~\ln(x )\leqslant{x^{\varphi>0}}$. Here I will use $\displaystyle \forall{x}>{0}~~\ln(x)\leqslant\sqrt{x}$. If you would like to prove this to yourself try finding the absolute maximum of $\displaystyle f(x)=\sqrt{x}-\ln(x)$ and noting its sign.

    So now here is what we have

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

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

    So now let us note something

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

    Cutting out the uneccessary step gives us

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

    So this implies that

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

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

    So we can see that

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

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


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

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