Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\limits_{x\to \infty}\frac{2\cdot 4^x-4^{2x}}{2\cdot 3^x-3^{2x}}=\infty$$
The limit should be $0$. Could someone point out what is wrong?

Use equivalents:


*

*$\ln(1+4^x)\sim_\infty\ln 4^x=x\ln 4$,

*similarly  $\ln(1+3^x)\sim_\infty\ln 3^x=x\ln 3$,


hence $$\frac{\ln(1+4^x)}{\ln(1+3^x)}\sim_\infty\frac{x\ln 4}{x\ln 3}=\frac{\ln 4}{\ln 3}.$$
