Why do I first need to bring $-4x$ into the numerator in $\lim_{x\to \infty} 4x^2/(x-2) - 4x$ I tried solving the question in the title as follows:
$$\lim_{x\to \infty} \frac{4x^2}{x-2} - 4x \to 4x - 4x \to 0$$
However, apparently that first step ($\to 4x - 4x$) was wrong, and I should first have brought the second $4x$ into the numerator.
My question is not how I need to solve the question, as I know that now. My question is why what I did was wrong, as I lack any intuition for it, and it seems a mystery to me.
 A: As $x \to \infty$, $\frac{4x^2}{x-2}$ and $4x$ are asymptotically equivalent. However, the notion of asymptotic equivalence is of relative equivalence (in the sense that their ratios tend to $1$). We cannot deduce anything about the differences. The differences may be fixed (e.g. $x^2 \sim x^2 + 1$), tend to $0$ (e.g. $x^2 \sim x^2 + \frac1x$) or tend to infinity (e.g. $x^2  \sim x^2 + x$). 
Therefore when working with limits which involve differences, the asymptotic equivalence becomes essentially irrelevant. This is why what you're doing is incorrect. 
A: What yo did was wrong for at least three reasons:
(1) Why would $\;\lim_{x\to\infty}\frac{4x^2}{x-2}=4x\;?$
(2) You can't take the limit when $\;x\to\text{whatever}\;$ and still remain with $\;x\;$ in the final expression
(3) You can't pass to the limit in only part of the expression, which is what ou did here: you let $\;4x\;$ untouched and pretended to calculate the first summand's limit (also this wrongly, as noted in (1)) .
A: Hint
You can use long division first and get $$\frac{4x^2}{x-2}=4 x+8+\frac{16}{x}+\cdots$$
The other way could be $$\frac{4x^2}{x-2}=4\frac{x^2}{x-2}=4\frac{x^2-4x+4+4x-4}{x-2}=4\frac{(x-2)^2+4(x-1)}{x-2}=4\left(x-2+4\frac{x-1}{x-2} \right)=4\left(x-2+4\frac{x-2+1}{x-2} \right)=4\left(x-2+4\left(1+\frac{1}{x-2}\right) \right)=$$ $$4\left(x+2+\frac 4{x-2}\right)=4x+8+\frac {16}{x-2}$$
A: In general if you have a set $A \subset \mathbb{R}$ and two functions $f , g : A \to \mathbb{R}$ with $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = \infty$ you can't say that $\lim_{x \to \infty} (f - g)(x) = 0$; for example, take $f(x) = x^2$ and $g(x) = x$.
A: You cannot write "$4x^2/(x-2)\to 4x$ as $x\to \infty$ " as it is gibberish: Check the definition of a limit.
In many cases $A(x)/B(x)-C(x)$ may have a limit when neither $A(x)/B(x)$ nor $C(x)$ does.
In your Q, it is worthwhile to put the expression into a form with a common denominator and see what you get. You get $8x/(x-2)$ which can be seen to be equal to $8+16/(x-2) ,$ which converges to $8$ as $x\to \infty.$
