Comparing the growth of two functions This is my almost final result for comparing two functions.
$$\lim\limits_{x\to \infty} =\frac{f(x)}{g(x)}=\frac{4x}{2x}=2$$
From here can we say which one of these functions grows faster?
 A: Assuming you mean:
$$\lim_{x\to \infty} \frac{f(x)}{g(x)}=\frac{4x}{2x}=2\tag1$$
without the "$=$" sign directly after $\lim$.
This reads a bit like a trick question...
The free variable $x$ in $\dfrac{4x}{2x}$ doesn't relate to the bound variable $x$ in the limit, so you could have written just as well $\dfrac{4z^4}{2z^4}$ or $\dfrac{4\cdot\mathrm{foo}}{2\cdot\mathrm{foo}}$.
Moreover, we just know the limit of the quotient of the two functions as $x\to\infty$, and $f$ and $g$ can be selected from an infinite set.  By no means it's implied that $f(x)=4x$ and $g(x)=2x$. Just to give some examples of functions that match $(1)$:
Example 1: Take
$$\begin{align}
f(x) &= -4 x^2 \\
g(x) &= -2 x^2
\end{align}$$
and $f$ drops faster than $g$.
Example 2: Take
$$\begin{align}
f(x) &= 4 + \frac1x\sin(x) \\
g(x) &= 2 - \frac1x\sin(x)
\end{align}$$
$f$ is decreasing each time $g$ is increasing.
Example 3: Take
$$\begin{align}
f(x) &= -4 \\
g(x) &= -2
\end{align}$$
$f$ and $g$ are not growing / decreasing at all, and $f$ is smaller than $g$.
etc. etc.
