How do I change the form of this case to apply L'hosptial's rule?
lim (ln|x-1| - ln|x+1|)
x->+inf
It seem to approach inf - inf, which is indefinite. In the cases I've seen this, they're fractions, and combining them does the job, but what about a case like this? Oh, and it apparently approaches 0.

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your concerned limit is :
\[\lim_{x\to \infty } \mid { \frac{x-1}{x+1} } \mid \]

forgot ln of | x-1/x+1|

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