Evaluation of limit using L'Hopital's rule:
As $x \to +\infty$, $\ln x \to +\infty$ and $x^2 \to +\infty$, giving an indeterminate form of type $\left[\frac{\infty}{\infty}\right]$.
Applying L'Hopital's rule by differentiating numerator and denominator with respect to $x$:
$\lim_{x \to +\infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x^2)} = \lim_{x \to +\infty} \frac{1/x}{2x} = \lim_{x \to +\infty} \frac{1}{2x^2}$.
As $x \to +\infty$, the denominator $2x^2 \to +\infty$, so:
$\frac{1}{2x^2} \to 0$.

Option breakdown:

• $1$ is incorrect.
• $-\infty$ is incorrect.
• $+\infty$ would occur if the power of $x$ were inverted.
• $0$ is the exact limit.