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.

Let $K>1$ be a constant. Let $y = K^x$. Then $\ln y = x \ln K$. So 
$\dfrac{y'}{y} = \ln K$. Hence $$\frac{d}{dx}K^x = \ln(K) K^x \tag 1$$.
As a reality check, note that this gives you $\frac{d}{dx}e^x = e^x$.
Using L'Hospital's rule, we see that 
$\displaystyle \lim_{x \to \infty}\dfrac{x}{K^x} =
               \lim_{x \to \infty}\dfrac{1}{\ln(K)K^x} =  0$. Using induction, we see that
$$\displaystyle \lim_{x \to \infty}\dfrac{x^n}{K^x} = 0 \tag 2$$
\begin{align}
   \lim_{x \to \infty}\frac{7x^4+x^2\, 3^x+2}{x^3+x\,4^x+1} 
   &= \lim_{x \to \infty}\frac
   {\dfrac{7x^3}{4^x} + \dfrac{x}{\left(\frac 43\right)^x} + \dfrac{2}{4^x}}
   {\dfrac{x^2}{4^x} +1 + \dfrac{1}{x 4^x}} = 0
\end{align}
