limit of ln(x) is negative infinity?

Question

I have a question on this specific question from my textbook.

How is limit of $\ln(x^2)$ as $x$ approaches $0$ equal to $-\infty$?

Shouldn't it be undefined?

Since this chapter is without l'Hospital's rule, I would like to know without using the rule.

Answer 1

Note that $\ln (x^2)$ is defined for any $x\in\mathbb R \setminus\{0\}$ and therefore the limit exists, indeed just take $y=x^2 \to 0^+$ then

$$\lim_{x\to 0} \ln (x^2)=\lim_{y\to 0^+} \ln y=-\infty$$

Answer 2

$|x|<e^{-\frac M 2}$ implies $x^{2} <e^{-M}$ which implies $\ln (x^{2}) <-M$. By definition of limit this gives $\lim_{ x \to 0} \ln (x^{2})=-\infty$.