Answer
$-\infty$
Work Step by Step
Given: $\lim\limits_{x\to0^+}\dfrac{\ln x}{x}$
Here, $\lim\limits_{x\to0^+}(\ln x)=-\infty$ (as $x\to0^+$, $\ln x\to-\infty$) and $\lim_{x\to0^+}x=0$ . It shows a form of $\frac{-\infty}{0}$, thus we cannot apply L'Hospital's Rule .
Since, $x\to0^+$, $\ln x\to-\infty$, that is, towards a negatively very large number, while $x\to0^+$,that is, towards a very small positive number.
we can conclude that the numerator part approaches negative infinity.and becomes even smaller.
Therefore, $\lim\limits_{x\to0^+}\dfrac{\ln x}{x}=-\infty$
