Find without using derivatives the following limit $$\lim_\limits{x\to +\infty}{\left[\frac{\ln\left(1+2^x\right)}{\ln\left(1+3^x\right)}\right]}$$
I have no idea what to do! I have tried squeezing it, using substitutions, but it is untouchable! Any hint?
On
The number $1$ is negligible faces to $2^x$ and $3^x$ as $x$ tends to infinity so
$$\ln(1+2^x)\sim \ln(2^x)=x\ln2$$ and similarly for the denominator so we find the limit $\frac{\ln2}{\ln3}$.
On
Notice, $$\lim_{x\to \infty}\frac{\ln(1+2^x)}{\ln(1+3^x)}$$ $$=\lim_{x\to \infty}\frac{\ln2^x(1+2^{-x})}{\ln3^x(1+3^{-x})}$$ $$=\lim_{x\to \infty}\frac{\ln2^x+\ln(1+2^{-x})}{\ln3^x+\ln(1+3^{-x})}$$ $$=\lim_{x\to \infty}\frac{x\ln2+\ln(1+2^{-x})}{x\ln3+\ln(1+3^{-x})}$$ $$=\lim_{x\to \infty}\frac{\ln2+\frac{1}{x}\ln(1+2^{-x})}{\ln3+\frac{1}{x}\ln(1+3^{-x})}$$ $$=\frac{\ln2+0}{\ln3+0}=\color{red}{\frac{\ln2}{\ln3}}$$
If one is in a squeezing mood, let our function be $f(x)$. For positive $x$ we have $$\frac{\ln(2^x)}{\ln(3^x+3^x)}\lt f(x)\lt \frac{\ln(2^x+2^x)}{\ln(3^x)}.$$