# 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?

• $x^2 - 4x = (x-2)^2 - 4$. The constant $4$ is indeed irrelevant when $x \to \infty$. Oct 2, 2013 at 23:48
• @DanielFischer Thank you! This explains where the correct answer of $-2$ comes from.
– bdr9
Oct 3, 2013 at 0:10

• Thank you! I understand now why the $4x$ cannot be ignored.
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 $$\lim_{t\to0^+}\frac{1-2t+o(t^2)-1}{t}=-2$$