i have a problem with limit calculation

Question

$\lim\limits_{x \to 0+} ⁡{x}^{(x^x-1)}=,$

i tried L'Hôpital but i really don't know how

i got to the second derivative

Answer 1

$$\lim_{x\rightarrow0^+}x^{x^x-1}=\lim_{x\rightarrow0^+}\left(1+x^x-1\right)^{\frac{1}{x^x-1}\cdot\frac{(x^x-1)^2}{x}}=e^{\lim\limits_{x\rightarrow0^+}\frac{(x^x-1)^2}{x}}=$$ $$=e^{\lim\limits_{x\rightarrow0^+}2(x^x-1)x^x(1+\ln{x})}=e^{2\lim\limits_{x\rightarrow0^+}\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot x\ln^2x}=e^{2\cdot1\cdot0}=1.$$

Answer 2

write $$e^{\lim_{x\to 0^+}(x^x-1)\ln(x)}$$ and use L'Hospital Hint: $$\lim_{x \to 0^+}(x^x-1)\log(x)=0$$