Find $\lim_\limits{x\to -\infty}{\frac{\ln\left(1+3^x\right)}{\ln\left(1+2^x\right)}}$

HINT:

$$\frac{z}{z+1} \le \log (1+z)\le z \tag 1$$

for $z>-1$. Then, use the squeeze theorem.

SPOILER ALERT: Scroll over the highlighted region to reveal the solution.

Using $(1)$, we have $$\frac{3^x}{2^x(1+3^x)}<\frac{\log (1+3^x)}{\log (1+2^x)}<\frac{3^x(1+2^x)}{2^x}\tag 2$$As $x\to -\infty$ both the left-hand and right-hand sides of $(2)$ approach $0$. Therefore, by the squeeze theorem, the limit of the sequence of interest goes to $0$.

NOTE 1:

We can prove the inequalities in $(1)$ using standard tools, which don't require the use of derivatives. We need only to note that

$$e^x\ge 1+x \tag 2$$

For $x>-1$, we take the log of both sides to reveal

$$\log(1+x)\le x$$

which is the right-hand side inequality of $(1)$. Likewise, if we make the substitution $x=-z/(z+1)$ in $(2)$, we obtain

$$e^{-z/(1+z)}\ge 1-\frac z{z+1}=\frac1{z+1}$$

whereupon rearranging yields for $z>-1$

$$\log(1+z)\ge \frac{z}{z+1}$$

as was to be shown.

NOTE 2:

We can prove the inequality given in $(2)$ using the limit definition of $e^x$

$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$$

I showed in THIS ANSWER, without the use of derivatives, that $\left(1+\frac xn\right)^n$ is a monotonically increasing function of $n$ (If $x<0$, this is true for $n>|x|$). Then, we have

$$e^x\ge \left(1+\frac xn\right)^n\ge 1+x$$

where this last inequality follows from Bernoulli's inequality. And we are done!

Notice, $$\lim_{x\to -\infty}\frac{\ln(1+3^x)}{\ln(1+2^x)}$$ substituting $x=-t$, we get
$$\lim_{t\to \infty}\frac{\ln(1+3^{-t})}{\ln(1+2^{-t})}$$ $$=\lim_{t\to \infty}\left(\frac{\frac{\ln(1+3^{-t})}{3^{-t}}}{\frac{\ln(1+2^{-t})}{2^{-t}}}\right)\times \frac{2^t}{3^t}$$

$$=\left(\frac{\lim_{t\to \infty}\frac{\ln(1+3^{-t})}{3^{-t}}}{\lim_{t\to \infty}\frac{\ln(1+2^{-t})}{2^{-t}}}\right)\times\lim_{t\to \infty} \left(\frac{2}{3}\right)^t$$ $$=\frac{1}{1}\times (0)=\color{red}{0}$$