Limits using L'Hopital's rule $\lim_{x\to0^+} (x^{x^x-1})$ Could you help me with this one? Thanks. The answer should be 1, somehow. I tried everything I know, but I couldn't solve it.
$$\lim_{x\to0^+} (x^{x^x-1})$$
 A: Write everything in an accurate way:
$$x^{x^x-1} = \exp ((\log x )(x^x-1)) = \exp \left( x\log^2 x\left(
\frac{e^{x \log x}-1}{x \log x}\right) \right)$$
Now, since $\lim_{h \to 0}\frac{e^h-1}{h} =1$ and $\lim_{x \to 0}x \log^2 x = 0$ you get that the whole limit is
$$\exp(0 \cdot 1) = e^0 = 1.$$
A: HINT:
$$\lim_{x\to 0}\left(x^{x^x-1}\right)=\lim_{x\to 0}\exp\left(\ln\left(x^{x^x-1}\right)\right)=\lim_{x\to 0}\exp\left(\left(x^x-1\right)\ln(x)\right)=\exp\left(\lim_{x\to 0}\left(x^x-1\right)\ln(x)\right)$$
A: APPROACH 1:
Although the OP requested an approach using L'Hospitals' Rule, I thought it might be instructive to present an approach that uses asymptotic analysis.  To  that end, we write for $x\to 0^+$
$$\begin{align}
x^{x^x-1}&=e^{\log(x)\,\left(x^x-1\right)}=e^{\log(x)\,\left(e^{\log(x)\,x}-1\right)}\\\\
&= e^{x\log^2(x)+O\left(x^2\log^3(x)\right)}
\end{align}$$
Inasmuch as both $\lim_{x\to 0^+}x\log^2(x)=0$ and $\lim_{x\to 0^+}x^2\log^3(x)=0$, we have
$$\lim_{x\to 0^+}x^{x^x-1}=1$$

APPROACH 2:
If one insists on using L'Hospital's Rule, we can proceed as follows.
$$\begin{align}
\lim_{x\to 0^+}\log(x)\,\left(e^{\log(x)\,x}-1\right)&=\lim_{x\to 0^+}\frac{\left(e^{\log(x)\,x}-1\right)}{1/\log(x)}\\\\
&=\lim_{x\to 0^+}\frac{(1+\log(x))\,e^{x\log(x)}}{-\frac{1}{x\log^2(x)}}\\\\
&=-\lim_{x\to 0^+}x\log^2(x)(1+\log(x))\,e^{x\log(x)}\\\\
&=0
\end{align}$$
Therefore, we recover the aforementioned result that used asymptotic analysis
$$\lim_{x\to 0^+}x^{x^x-1}=1$$
as expected!
A: $\lim_{x \to 0}\ln(x^{(x^x-1)})= \lim_{x \to 0}(x^x-1)\ln(x)=\lim_{x \to 0} \frac{(x^x-1)}{x} \frac{-\ln(1/x)}{1/x}$
Applying L hospital rule to first term 
$=\lim_{x \to 0} x^x(\ln(x)+1)\frac{-\ln(1/x)}{1/x}= \lim_{x \to 0} x^x(-\ln(1/x)+1)\frac{-\ln(1/x)}{1/x}= \lim_{x \to 0} x^x\frac{(\ln(1/x)^2-\ln(1/x))}{1/x}$
Note that, $\lim_{x \to 0} x^x=1$ ,  $\ln(1/x)/(1/x)=0$ , $(\ln(1/x))^2/(1/x)=0$
$\lim_{x \to 0} x^x\frac{(\ln(1/x)^2-\ln(1/x))}{1/x}= 1(0-0)=0$
$\lim_{x \to 0} (x^{(x^x-1)})= \exp(0)=1$
