1. ## Limit

Evaluate:
$
\lim_{x\to 0+}(x^{x^x}-x^x)
$

2. Originally Posted by pankaj
Evaluate:
$
\lim_{x\to 0+}(x^{x^x}-x^x)
$
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 $0-1=-1$