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