TITLE: How to calculate $\lim_{x \to -\infty}\ln(x^2 - 1)$?
QUESTION [0 upvotes]: How can I show that 
$$ \lim_{x \to -\infty} \ln(x^2 - 1) = \infty$$
My approach would be to compute the derivative and simply show that the derivative is always positive for large inputs of type $x^2 - 1$, so log is an increasing function and goes to infinity as $x^2 -1$ goes to infinity.
Another approach seems to get me into some trouble, i.e. trying to using log rules:
$$ \lim_{x \to -\infty} \ln(x^2 - 1)$$
$$ = \lim_{x \to -\infty} \ln( (x-1) \cdot (x+1))$$
$$ = \lim_{x \to -\infty} \ln(x-1) + \ln(x+1),$$
which now leaves me stuck and can't evaluate these limits, as x goes to negative infinity.  How could I resolve this?
Thanks,

REPLY [0 votes]: $f(x)=x^2=(-x)^2= f(-x)$ is even.
$\lim_{x \rightarrow - \infty} x^2 = \lim_{x \rightarrow \infty}x^2$
Let $y:= x^2-1>0.$
Consider:
$\lim_{y \rightarrow \infty} \log (y) =\infty.$
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