How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$? 
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$ 

I can't  seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.

Numerator : $7x^4 + x^2 3^x + 2 \lt 3x^2 3^x$
for $x\gt 4$ (say).
Denominator: $x^3 + x4^x +1 \gt x4^x$ 
for $x \gt 0.$
$0 \lt \dfrac{7x^4 + x^2 3^x +3}{x^3 +4^x +1} \lt $
$ \dfrac{3x^23^x}{x4^x} = 3x (\dfrac{3}{4})^x =
3x(\dfrac{ 4}{3})^{-x} $.
Note: $y := (\dfrac{4}{3})^{-x} = e^{-x \log(4/3)}$.
Let $a:= \log(4/3) \gt 0.$
Using : 
$\star) \lim_{x \rightarrow \infty} x^k e^{-x} = 0$, $ k \in \mathbb{N}$:
$z:= ax$;
$3xe^{-ax} = (3/a) (ax)e^{-ax} = (3/a)ze^{-z}$.
Altogether :
$0\le \lim_{x \rightarrow \infty} \dfrac{7x^4 +3x^2 x^3 +2}{x^3 + x4^x + 1} $
$\lt \lim_{z \rightarrow \infty}(3/a)ze^{-z} = 0$.
