@topper 1. correct
2. Well, it's obviously true. But if you evaluate $ln(x)$ as $-\infty$, then $ln^2(x)$ is more 'like' $\infty^2$ in the sense of how *strong* it goes there. But then again, it's not really so, it's just to give you a sense of why evaluating it as you did still makes sense with the result of $\infty$