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

I am trying to intuitively understand why the solution to the following problem is $-2$. $$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$ $$\lim_{x\to\infty}(\sqrt{x^2-4x}-x)\frac{\sqrt{x^2-4x}+x}{\sqrt{x^2-4x}+x}$$ $$\lim_{x\to\infty}\frac{x^2-4x-x^2}{\sqrt{x^2-4x}+x}$$ $$\lim_{x\to\infty}\frac{-4x}{\sqrt{x^2-4x}+x}$$ $$\lim_{x\to\infty}\frac{-4}{\sqrt{1-\frac{4}{x}}+1}$$ $$\frac{-4}{\sqrt{1-0}+1}$$ $$\frac{-4}{2}$$ $$-2$$ I can understand the process that results in the answer being $-2$. However, I expected the result to be $0$. I have learned that when dealing with a limit approaching $\infty$, only the highest degree term matters because the others will not be as significant. For this reason, I thought that the $4x$ would be ignored, resulting in: $$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$ $$\lim_{x\to\infty}\sqrt{x^2}-x$$ $$\lim_{x\to\infty}x-x$$ $$\lim_{x\to\infty}0$$ $$0$$ Why is the above process incorrect?

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$x^2 - 4x = (x-2)^2 - 4$. The constant $4$ is indeed irrelevant when $x \to \infty$. –  Daniel Fischer Oct 2 '13 at 23:48
@DanielFischer Thank you! This explains where the correct answer of $-2$ comes from. –  bdr9 Oct 3 '13 at 0:10

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.

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Thank you! I understand now why the $4x$ cannot be ignored. –  bdr9 Oct 3 '13 at 0:10
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 $$\lim_{t\to0^+}\frac{1-2t+o(t^2)-1}{t}=-2$$