[SOLVED] Why is limit of natural log equal to ln of limit?

**x3bnm** · November 16th 2009, 03:41 PM

Why is that

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

Can anyone kindly explain it?

**Drexel28** · November 16th 2009, 03:44 PM

Originally Posted by x3bnm

Why is that

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

Can anyone kindly explain it?

It isn't. The first one is 1 the second is 0[/tex]

**x3bnm** · November 16th 2009, 04:37 PM

Sorry the expression was wrong in my last post. Edited one is :

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

I found them Thomas' Calculus solution book as part of solving a integral(Chapter 8.8 exercise9). But why are they equal?

**hjortur** · November 16th 2009, 04:44 PM

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

**x3bnm** · November 16th 2009, 04:53 PM

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

Can you elaborate on this?

**RockHard** · November 16th 2009, 04:57 PM

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

**x3bnm** · November 16th 2009, 05:05 PM

Thanks hjortur and RockHard. I get it now.

**redsoxfan325** · November 16th 2009, 05:07 PM

One of the definitions/implications of continuity is that if you have a sequence $\{x_n\}$ such that $\lim_{n\to\infty}x_n=x$ , then $\lim_{n\to\infty}f(x_n)=f(x)$

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