Limit of $\dfrac{(1+4^x)}{(1+3^x)}$? I don't remember how to find the limit in this case. I take $x$ towards $+\infty$.

$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$

I do not know where to start. I would instinctively say that $1$ can't be right because $4^x$ goes faster than $3^x$ and thus one would move towards infinity, but this is apparently not the case....

$$\lim_{x\rightarrow \infty}\dfrac{(1+4^x)}{(1+3^x)}=$$
$$\lim_{x\rightarrow \infty}\frac{1}{1+3^x}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\frac{1}{\lim_{x\rightarrow \infty} 1+3^x}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\frac{1}{\infty}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$0+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{3^{-x}+1}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{0+1}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{1}=$$
$$\lim_{x\rightarrow \infty}\left(\frac{4}{3}\right)^x=$$
$$\left(\frac{4}{3}\right)^{\lim_{x\rightarrow \infty}x}=\left(\frac{4}{3}\right)^{\infty}=\infty$$
