1. The problem is to find the limit of the function $$\frac{\ln x}{x^2}$$ as $$x$$ approaches $$0^+$$ (from the right side). 2. We want to evaluate $$\lim_{x \to 0^+} \frac{\ln x}{x^2}$$. 3. Recall that as $$x \to 0^+$$, $$\ln x \to -\infty$$ and $$x^2 \to 0^+$$. 4. This is an indeterminate form of type $$\frac{-\infty}{0^+}$$ which suggests the limit might be $$-\infty$$ or diverge. 5. To analyze the limit, substitute $$x = \frac{1}{t}$$ so that as $$x \to 0^+$$, $$t \to +\infty$$. 6. Rewrite the limit in terms of $$t$$: $$\lim_{x \to 0^+} \frac{\ln x}{x^2} = \lim_{t \to +\infty} \frac{\ln \left( \frac{1}{t} \right)}{\left( \frac{1}{t} \right)^2} = \lim_{t \to +\infty} \frac{-\ln t}{\frac{1}{t^2}} = \lim_{t \to +\infty} -\ln t \cdot t^2$$ 7. As $$t \to +\infty$$, $$\ln t$$ grows slowly but $$t^2$$ grows much faster, so $$-\ln t \cdot t^2 \to -\infty$$. 8. Therefore, the original limit is: $$\boxed{-\infty}$$ This means the function $$\frac{\ln x}{x^2}$$ decreases without bound as $$x$$ approaches $$0$$ from the right side.