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.
 A: $$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
The most intuitive way  to solve this limit is to understand how the graphs of $b^x, b>0$, and how graphs of $x^2, x^3,x^4,\dots$ look. 
When you reach higher and higher values of $x$, $9999999^2\times3^{9999999}(x^23^x)$ is much bigger then $9999^4(7x^4)$, and $9999999\times 4^{9999999}(x4^x)$ is much bigger than $9999999^3(x^3)$.
Therefore, the terms $x^23^x$, and $x4^x$ "dominate" as $x\to\infty$, and so we can basically "simplify" this as follows:
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}=\lim_{x\to\infty}\frac{x^23^x}{x4^x}=\lim_{x\to\infty}x(\frac{3}{4})^x$$
It is also a common fact that $\lim_{x\to\infty}(\frac{a}{b})^x$, where $a<b$, this limit is $0$. (helps to know this graph too)
Therefore, we see that the $(\frac{3}{4})^x$ term goes to $0$, or a very small number. Thus, $x$, which is massive, times a really small number, is also a very very small number, and as you can probably guess, this number tends to get closer and closer to $0$.
The reason I tell you like this and not with math itself is because the math is a little tricky with $e$ and $\ln$ and if you can understand it this way, it will really help your life in the future.
This way you don't need to "get rid" of anything. 
A: Use equivalents:
$$\begin{aligned}
7x^4+x^2\, 3^x+2&\sim_\infty x^2 3^x\\
x^3+x\,4^x+1&\sim_\infty x\,4^x
\end{aligned}\quad\text{hence}\quad\frac{7x^4+x^2\, 3^x+2}{x^3+x\,4^x+1}\sim_\infty\frac{ x^2 3^x}{ x\,4^x}=x\Bigl(\frac34\Bigr)^{\!x}\to 0.$$
A: hint 
factor out by $x^23^x $ in numerator to get
$$x^23^x \Bigl (1+\frac {2}{x^23^x}+7e^{x (2\frac {\ln (x)}{x}-\ln (3))} \Bigr)$$
for denominator, factor out by $x4^x$ as
$$x4^x\Bigl ( 1+\frac {1}{x4^x}+ e^{x (2\frac {\ln (x)}{x}-\ln (4))} \Bigr) $$
The limit will be
$$\lim_{\infty}xe^{x (\ln (3)-\ln (4))}=$$
$$\lim_{\infty}e^{x(\ln (3)-\ln (4)+\frac {\ln (x)}{x})}=0$$
A: Divide the numerator and denominator by $x4^{x}$ and then it is clear from the standard limit $$\lim_{x\to\infty} \frac{x^{n}}{a^{x}} = 0, a > 1\tag{1}$$ that the numerator tends to $0$ and denominator tends to $1$. Thus the desired limit is $0$.
One may ask: why divide by $x4^{x} $? Because that is the term which grows to $\infty$ much faster than any other term. To be precise, the ratio of any other term and this term tends to $0$ via the equation $(1)$ above. 
A: 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}
A: 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} $
$\lt \lim_{z \rightarrow \infty}(3/a)ze^{-z} = 0$.
