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....
 A: we will use the fact that for $r > 1, \,\lim_{x \to \infty} r^x = \infty.$
you have $$\frac{1+4^x}{1+3^x} \ge \frac{4^x}{3^x +3^x} = \frac12\left(\frac43\right)^x \to \infty \text{ as } x \to \infty.$$
A: Divide the numerator and denominator by $3^x$ and let x go to $+\infty$.
Then you will easily see what the limit is. You get:
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x} = \lim\limits_{x\to \infty} \dfrac{(\frac{1}{3})^x+(\frac{4}{3})^x}{(\frac{1}{3})^x+1^x} = \frac{0 + (+\infty)}{0+1} = +\infty$.    
A: 
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$

$=\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}$+
$\lim\limits_{x\rightarrow \infty}\frac{4^x}{1+3^x}$
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 $\infty$
$=\lim\limits_{x\rightarrow \infty}(\frac{4}{3})^x$
$=\left(\frac{4}{3}\right)^{\lim_{x\rightarrow \infty}x}=\left(\frac{4}{3}\right)^{\infty}=\color{red}\infty$
A: $$\lim_{x\rightarrow \infty}\dfrac{(1+4^x)}{(1+3^x)}=$$
$$\lim_{x\rightarrow \infty}\frac{1}{1+3^x}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\frac{1}{\lim_{x\rightarrow \infty} 1+3^x}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\frac{1}{\infty}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$0+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{3^{-x}+1}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{0+1}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{1}=$$
$$\lim_{x\rightarrow \infty}\left(\frac{4}{3}\right)^x=$$
$$\left(\frac{4}{3}\right)^{\lim_{x\rightarrow \infty}x}=\left(\frac{4}{3}\right)^{\infty}=\infty$$
A: We have :
$\frac{1+4^x}{1+3^x} = \frac{1+\exp(x(\ln 4))}{1+\exp(x(\ln 3))} = \exp(x(\ln (4/3)))\times \frac{1+\exp(-x(\ln4))}{1+\exp(-x\ln3))}$
So $\lim \limits_{x\to \infty} \frac{1+4^x}{1+3^x} = \infty \times 1 = \infty$
