Answer
$1$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0^{+}} x^x$
Let us consider that $y=\lim\limits_{x \to 0^{+}} x^x \implies \ln y=\lim\limits_{x \to 0^{+}} [x \ln (x)]$
Apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
$ \ln y=\lim\limits_{x \to 0^{+}}\dfrac{-x^2}{x}\\=\lim\limits_{x \to 0^{+}} (-x) \\=0$
Thus, we have: $\lim\limits_{x \to 0^{+}} y=e^0=1$
