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