Question

find the limit of (ln(ax-1)-ln(bx+1)) as x approaches infinity

Answer 1

if you plug infinity in for x you will get infinity minus infinity which is indeterminate so you need to find a way to get into the form 0/0 and use l'hopital's rule. That link from your last question should help you.

Answer 2

I just don't understand what the question is asking.

Answer 3

could you possibly explain the process to me?

Answer 4

have you ever solved limit question before ?

Answer 5

Yes, however I still do understand how to answer the problem.

Answer 6

\[\lim_{x\rightarrow \infty} \ln(ax-1) - \ln(bx+1)\]By log property: \(\log m -\log n = \log (\frac{m}{n})\), we get
\[=\lim_{x\rightarrow \infty} \ln(\frac{ax-1 }{bx+1})\]Divide both numerator and denominator by x, we get:
\[=\lim_{x\rightarrow \infty} \ln(\frac{a-\frac{1}{x} }{b+\frac{1}{x}})\]Do you have any problem about the above steps? Can you do it from here?