limits

**sss** · December 1st 2008, 06:05 PM

how to find lim (-ln(x))^x
x->0+

**Mathstud28** · December 1st 2008, 08:27 PM

Originally Posted by sss

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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