Evaluate:
$\displaystyle
\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 $\displaystyle \lim_{x\to{0}}x^x$
Way One: if we let $\displaystyle x=\frac{1}{\varphi}$ this limit transforms to $\displaystyle \lim_{\varphi\to\infty}\left(\frac{1}{\varphi}\rig ht)^{\frac{1}{\varphi}}$, now by the connection of the Root and Ratio tests, this is equivalent to $\displaystyle \lim_{\varphi\to\infty}\frac{\frac{1}{\varphi+1}}{ \frac{1}{\varphi}}=1$
Way Two: After the sub $\displaystyle \frac{1}{\varphi}\left(\frac{1}{\varphi}\right)^{\ frac{1}{\varphi}}$ we may rewrite it as $\displaystyle \lim_{\varphi\to\infty}e^{\ln\left(\left(\frac{1}{ \varphi}\right)^{\frac{1}{\varphi}}\right)}=\lim_{ \varphi\to\infty}e^{\frac{-\ln(\varphi)}{\varphi}}$ and by the continuity of the exponentiat function across its entire domain we may rewrite this as $\displaystyle e^{\lim_{\varphi\to\infty}\frac{-\ln(\varphi)}{\varphi}}$. Now by any number of methods (I leave this to you) $\displaystyle \lim_{\varphi\to\infty}\frac{\ln(\varphi)}{\varphi }=0$ so we have our limit is equal to $\displaystyle e^0=1$
Now my fingers are getting tired so lets only consider one method for this one. Once again let us rewrite our limit as $\displaystyle \lim_{x\to{0^+}}e^{\ln\left(x^{x^x}\right)}=e^{\li m_{x\to{0}}x^x\cdot\ln(x)}$
Now $\displaystyle x^x\to{1}$ and $\displaystyle \ln(x)\to-\infty$
So we have that $\displaystyle x^x\cdot\ln(x)\to-\infty$ and that makes $\displaystyle e^{\lim_{x\to{0^+}}x^x\ln(x)}\to{0}$
So the asnwer to your actual limit is $\displaystyle 0-1=-1$