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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

3 Answers 3

up vote 5 down vote accepted

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 $$ \lim_{t\to0^+}\frac{1-2t+o(t^2)-1}{t}=-2 $$

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Neglecting lower order terms while dealing with certain limits is a sort of "thumb rule" which applies in certain circumstances and its proper justification in those circumstances is based on the standard rules of limits. You are trying to assume that this thumb rule is more fundamental than the rules of limits and can be used in any circumstances. After gaining reasonable experience with limits an expert is able to figure out where and how to apply this thumb rule correctly. But for a beginner it makes sense to apply the rules of limits rather than rely on these non-fool-proof thumb rules.

If you have a look at the comment from Daniel Fischer you can see how he is able to use the thumb rule effectively without any problem by transforming the expression under square root properly and then ignoring the constant term. The fundamental issue in applying the thumb rule is to figure out which terms to ignore and which ones to keep.

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