limit of ln(x) is negative infinity? [closed]
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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.
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$\begingroup$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
$$\lim_{x\to 0} \ln (x^2)=\lim_{y\to 0^+} \ln y=-\infty$$
$\endgroup$ 3 $\begingroup$$|x|<e^{-\frac M 2}$ implies $x^{2} <e^{-M}$ which implies $\ln (x^{2}) <-M$. By definition of limit this gives $\lim_{ x \to 0} \ln (x^{2})=-\infty$.
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