TITLE: Find $\lim_\limits{x\to -\infty}{\frac{\ln\left(1+3^x\right)}{\ln\left(1+2^x\right)}}$
QUESTION [8 upvotes]: Prove the following limit without using approximations and derivatives: $$\lim_\limits{x\to -\infty}{\frac{\ln\left(1+3^{x}\right)}{\ln\left(1+2^{x}\right)}}=0$$

I cannot think of any possible factorization or inequality (so that I could use the Squeeze Theorem) that doesn't use derivatives so as to find this limit. Any hint?

REPLY [7 votes]: Notice, $$\lim_{x\to -\infty}\frac{\ln(1+3^x)}{\ln(1+2^x)}$$
substituting $x=-t$, we get
$$\lim_{t\to \infty}\frac{\ln(1+3^{-t})}{\ln(1+2^{-t})}$$
$$=\lim_{t\to \infty}\left(\frac{\frac{\ln(1+3^{-t})}{3^{-t}}}{\frac{\ln(1+2^{-t})}{2^{-t}}}\right)\times \frac{2^t}{3^t}$$
$$=\left(\frac{\lim_{t\to \infty}\frac{\ln(1+3^{-t})}{3^{-t}}}{\lim_{t\to \infty}\frac{\ln(1+2^{-t})}{2^{-t}}}\right)\times\lim_{t\to \infty} \left(\frac{2}{3}\right)^t$$
$$=\frac{1}{1}\times (0)=\color{red}{0}$$