Solving limits with ln without L’Hôpital’s rule

Question

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

Answer 1

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 $b$.

Answer 2

Use equivalents:

• If $a, b>0$, $\ln(1+\mathrm e^{ax})\sim_\infty\ln(\mathrm e^{ax})=ax$. Similarly $\ln(1+\mathrm e^{bx})\sim_\infty bx$. Hence $$\frac{\ln(1+\mathrm e^{ax})}{\ln(1+\mathrm e^{bx})}\sim_\infty\frac{ax}{bx}=\frac ab.$$
• If $a<0, b>0$, $\ln(1+\mathrm e^{ax})\downarrow \ln 1=0=$. Hence $\ln(1+\mathrm e^{ax})=o(1)$, so $$\frac{\ln(1+\mathrm e^{ax})}{\ln(1+\mathrm e^{bx})}=o\Bigl(\frac1{bx}\Bigr)\downarrow 0.$$
• If $a>0, b<0$, $\;\dfrac{\ln(1+\mathrm e^{bx})}{\ln(1+\mathrm e^{ax})}\downarrow 0$ by the previous case, hence $$\frac{\ln(1+\mathrm e^{ax})}{\ln(1+\mathrm e^{bx})}\to+\infty.$$
• If $a, b<0$, $\ln(1+\mathrm e^{ax})\sim_\infty\mathrm e^{ax}$. Similarly for $\;\ln(1+\mathrm e^{bx})$, hence $$\frac{\ln(1+\mathrm e^{ax})}{\ln(1+\mathrm e^{bx})}\sim_\infty\frac{\mathrm e^{ax}}{\mathrm e^{bx}}=\mathrm e^{(a-b)x}\to\begin{cases}+\infty&\text{if }\; a>b,\\0&\text{if }\; a<b,\\1&\text{if }\;a=b.\end{cases}$$

Answer 3

For $a,b>0$ and sufficiently large $x$ (actually for $x\geq 0$) you have $$ax = \ln e^{ax} \leq \ln(1+e^{ax}) \leq \ln(e^{ax}+e^{ax}) = \ln2+ax. $$ Now by applying this to the original limit, we get

$$ \frac{1}{\frac{\ln2}{ax}+\frac{b}{a}} = \frac{ax}{\ln2+bx} \leq \frac{\ln(1+e^{ax})}{\ln(1+e^{bx})} \leq \frac{\ln2+ax}{bx} = \frac{\ln2}{bx} + \frac{a}{b} $$

You can see that both surrounding functions approach $a/b$ for $x\to \infty$, so the original function does too by the Squeeze Theorem.