<s>[INST] 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.
 [/INST] 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 $$\frac{4x^2}{x-2}=4\frac{x^2}{x-2}=4\frac{x^2-4x+4+4x-4}{x-2}=4\frac{(x-2)^2+4(x-1)}{x-2}=4\left(x-2+4\frac{x-1}{x-2} \right)=4\left(x-2+4\frac{x-2+1}{x-2} \right)=4\left(x-2+4\left(1+\frac{1}{x-2}\right) \right)=$$ $$4\left(x+2+\frac 4{x-2}\right)=4x+8+\frac {16}{x-2}$$
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