I have a question on my maths assignment for this week that is stumping me. It goes like this
$$
\lim_{x\to\infty}\frac{\ln(1+e^{ax})}{\ln(1+e^{bx})}
$$
Anyway, we are not allowed to use L’Hôpital’s rule. My professor gave the following hints:
take out the factor $e^{ax}$ and $e^{bx}$ of the arguments of the logarithms and use algebraic rules of logarithms.
I think my main problem is i'm not sure how to manipulate the ln. You can't make it into one ln as far as I know, and you can't simplify it any more. My main question is: how can I take out the factors of $e$?
edit: only for the cases a>0 and b>0
Best Answer
Hint: $$ \ln(1+e^{ax})=\ln(e^{ax}(e^{-ax}+1))=ax+\ln(e^{-ax}+1) $$ If $a>0$, then $\lim_{x\to\infty}e^{-ax}=0$.
However, you have to distinguishing between the cases $a>0$, $a=0$, $a<0$ and the same for $b$.