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?
 A: 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 $$\frac{\ln(1+4^x)}{\ln(1+3^x)}\sim_\infty\frac{x\ln 4}{x\ln 3}=\frac{\ln 4}{\ln 3}.$$
A: $$\lim_{x\to\infty}\space\frac{\ln(1+4^x)}{\ln(1+3^x)}=$$
$$\lim_{x\to\infty}\space\frac{\ln(1+4^x)}{\ln(3^x)+\ln(1+3^{-x})}=$$
$$\lim_{x\to\infty}\space\frac{\ln(1+4^x)}{\ln(3^x)}=$$
$$\lim_{x\to\infty}\space\frac{\ln(4^x)\ln(1+4^{-x})}{\ln(3^x)}=$$
$$\lim_{x\to\infty}\space\frac{\ln(4^x)}{\ln(3^x)}=$$
$$\lim_{x\to\infty}\space\frac{x\ln(4)}{x\ln(3)}=$$
$$\lim_{x\to\infty}\space\frac{\ln(4)}{\ln(3)}=\frac{\ln(4)}{\ln(3)}$$
A: Since it is of the form $\frac{\infty}{\infty}$ L'Hopital's rule gives \begin{align}\lim_{x\to \infty}\frac{(\ln{(1+4^x)})'}{(\ln{(1+3^x)})'}&=\frac{\ln 4}{\ln 3}\cdot\lim_{x\to \infty}\frac{4^x(1+3^x)}{3^x(1+4^x)}=\\&=\frac{\ln4}{\ln3}\cdot\lim_{x\to \infty}\frac{12^x\left(\frac1{3^x}+1\right)}{12^x\left(\frac{1}{4^x}+1\right)}=\frac{\ln4}{\ln3}\cdot\lim_{x\to \infty}\frac{\frac1{3^x}+1}{\frac{1}{4^x}+1}=\frac{\ln4}{\ln3}\cdot \frac11=\frac{\ln4}{\ln3}\end{align}
A: Note that the series $\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,x^n$ is valid only for $-1<x\le 1$.  We can write for $|x|>1$, 
$$\begin{align}
\log(1+x)&=\log x+\log \left(1+\frac1x\right) \\\\
&=\log x +\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,x^{-n} \tag 1
\end{align}$$  
Using $(1)$, we can write for $a>1$ 
$$\begin{align}
\log(1+a^x)&=\log(a^x)+\log(1+a^{-x})\\\\
&=x\log a+O(a^{-x}) \tag 2
\end{align}$$
From $(2)$ we can write
$$\begin{align}
\frac{\log(1+4^x)}{\log(1+3^x)}&=\frac{x\log 4+O(4^{-x})}{x\log 3+O(3^{-x})}\\\\
&=\frac{\log 3+O\left(\frac{4^{-x}}{x}\right)}{\log 4+O\left(\frac{3^{-x}}{x}\right)}
\end{align}$$
The limit as $x\to \infty$ is evident now as the numerator  tends to $\log 4$ and the denominator tends to $\log 3$.  Therefore, we obtain
$$\lim_{x\to \infty}\frac{\log(1+4^x)}{\log(1+3^x)}=\frac{\log 4}{\log 3}$$
