I'm having a hard time understanding why this limits equals to 0.
By simply using logarithm identity I get that this limit equals
$ \lim_{x\to 0} \ e^{x\ln x^{x}}=\lim_{x\to 0}\ e^{x^{2}\ln x}$ and by L'Hopital we get that the limit of the exponent is 0 and because $f(a)=e^{a}$ is continuous I get that the final limit should be 1.
I'd love your help with understanding what I did wrong.
Thank you.