<s>[INST] 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....
 [/INST] 
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$

$=\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}$+
$\lim\limits_{x\rightarrow \infty}\frac{4^x}{1+3^x}$
since $\lim\limits_{x\rightarrow \infty}\ (1+3^x)=\infty$
$\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}=0$
The leadind term in the denominator of $\frac{4^x}{1+3^x}$ is $3^x.$ Divide the numerator and denomirator by this:
$\lim\limits_{x\rightarrow \infty}\frac{(\frac{4}{3})^x}{1+3^{-x}}$
The expression $3^{-x}$ tends to zero as $x$ approches $\infty$
$=\lim\limits_{x\rightarrow \infty}(\frac{4}{3})^x$
$=\left(\frac{4}{3}\right)^{\lim_{x\rightarrow \infty}x}=\left(\frac{4}{3}\right)^{\infty}=\color{red}\infty$
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