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?
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?
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