This is what we had discussed in class
$$x^x=e^{x \ln x}$$ $$\lim\limits_{x\rightarrow{{0}^{+}}}x^x=\lim\limits_{x\rightarrow{{0}^{+}}}e^{x \ln x}$$
Now after this my teacher applied L'Hopital's and said the answer should be $1$, but won't that approach be incorrect since $\ln x$ is undefined at $0$ and this is not a $\frac{0}{0}$ form? $\ln x$ is only zero when $x=1$. Am I in missing something, or is my teacher's method is wrong?
I have tried this myself by another method. As $x\rightarrow{{0}^{+}}$ $\ln x$ becomes more negative and $x$ becomes smaller. So the answer should be $e^0=1$. Is there any way I can formalise this method using a more rigorous approach?
