Explanation for $\lim_{x\to\infty}\sqrt{x^2-4x}-x=-2$ and not $0$

Intuitively, the thing you want to look at with your second attempt is that it's infinity minus infinity. Because the highest order terms are on the same order of magnitude and cancel exactly, the lower-order terms are indeed important to determine what the limit will be.

Change the variable: set $t=1/x$, so you want to compute $$ \lim_{t\to0^+}\sqrt{\frac{1}{t^2}-\frac{4}{t}}-\frac{1}{t} = \lim_{t\to0^+}\sqrt{\frac{1-4t}{t^2}}-\frac{1}{t} = \lim_{t\to0^+}\frac{\sqrt{1-4t}-1}{t} $$ Now it should be clearer why the limit can't be $0$. The square root can be written $$ \sqrt{1-4t}=1+\frac{1}{2}(-4t)+o(t^2) $$ so the limit becomes $$ \lim_{t\to0^+}\frac{1-2t+o(t^2)-1}{t}=-2 $$