Calculating of $\lim_{x\to 0^+} x^{(x^x-1)}$ When I  want calculate $\lim_{x\to 0^+} x^{(x^x-1)}$ , I have to calculate $\lim_{x\to 0^+}(x^x-1)\ln x$ . I calculat $\lim_{x\to 0^+}x^x=\lim_{x\to 0^+}e^{x\ln x}=1$ so $\lim_{x\to 0^+}(x^x-1)\ln x$ is $0\times \infty$ . 
How can I calculate $\lim_{x\to 0^+}(x^x-1)\ln x$? 

$$\lim_{x\to 0^+} \exp((x^x-1)\ln x) =$$
$$\exp(\lim_{x\to 0^+} (x^x-1)\ln x) = \exp(\lim_{x\to 0^+} \frac{x^x-1}{\frac1{\ln x}}) = \exp(\lim_{x\to 0^+} \frac{\exp(x\ln x)-1}{\frac1{\ln x}})$$
Using Hopital
$$\exp(\lim_{x\to 0^+} \frac{\exp(x\ln x)-1}{\frac1{\ln x}})=\exp(\lim_{x\to 0^+} \frac{(\ln x+1)\exp(x\ln x)}{\frac{-1}{x \ln^2 x}})=$$
$$=\exp(-\lim_{x\to 0^+} x \ln^2 x(\ln x+1)\exp(x\ln x))$$
Here we have $x \ln^3x$ and $x$ overcomes over $\ln^3 x$ the result is $\exp(0)=1$
