<s>[INST] Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ = 
 $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \ln(1+\frac{1}{2^x})}{\ln3^x + \ln(1+\frac{1}{3^x})}\right)$ = 
 $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \frac{1}{2^x}\ln(1+\frac{1}{2^x})^{2^{x}}}{\ln3^x + \frac{1}{3^x}\ln(1+\frac{1}{3^x})^{3^{x}}}\right)$ = $\cdots$
This method is not working.
 [/INST] If we use the well known  limit
$$\lim_{X\to 0}\frac{\ln(1+X)}{X}=1$$
with $X=2^x\;$ and $\;X=3^x,\;\;$we find
$$\lim_{x\to-\infty}\left( \frac{2}{3} \right)^x=+\infty$$
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