Why is that

$\displaystyle \lim _{x\to -\infty} \frac{x+1}{x-1} = \ln(\lim _{x\to -\infty} \frac{x+1}{x-1})$

Can anyone kindly explain it?

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- Nov 16th 2009, 03:41 PMx3bnm[SOLVED] Why is limit of natural log equal to ln of limit?
Why is that

$\displaystyle \lim _{x\to -\infty} \frac{x+1}{x-1} = \ln(\lim _{x\to -\infty} \frac{x+1}{x-1})$

Can anyone kindly explain it? - Nov 16th 2009, 03:44 PMDrexel28
- Nov 16th 2009, 04:37 PMx3bnm
Sorry the expression was wrong in my last post. Edited one is :

$\displaystyle \lim _{b\to -\infty} (\ln(|\frac{b-1}{b+1}|) ) = \ln(\lim _{b\to -\infty} \frac{b-1}{b+1}) $

I found them Thomas' Calculus solution book as part of solving a integral(Chapter 8.8 exercise9). But why are they equal? - Nov 16th 2009, 04:44 PMhjortur
Because ln(x) is continous for every $\displaystyle x>0$

- Nov 16th 2009, 04:53 PMx3bnmQuote:

Because ln(x) is continous for every $\displaystyle x > 0$

- Nov 16th 2009, 04:57 PMRockHard
The function is not discontinuous on any interval greater than 0, or does not have any asymptotic holes on intervals greater than 0, simply look at the graph as x is > 0 and it continues to reach infinity in terms of limits

Finding the graph is simple, just use a calculator or manually find the "sequence of terms" for each x, like x = 1, x = 2, x = 3...and so on to make the graph and youll see why - Nov 16th 2009, 05:05 PMx3bnm
Thanks hjortur and RockHard. I get it now.(Happy)

- Nov 16th 2009, 05:07 PMredsoxfan325
One of the definitions/implications of continuity is that if you have a sequence $\displaystyle \{x_n\}$ such that $\displaystyle \lim_{n\to\infty}x_n=x$, then $\displaystyle \lim_{n\to\infty}f(x_n)=f(x)$