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(I) $$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$ $$\lim_{x \to \infty } \, \left(x\sqrt{1+1/x}-x\sqrt{1-1/x}\right)=$$ $$\lim_{x \to \infty } \, \left(x\sqrt{1}-x\sqrt{1}\right)=\lim_{x \to \infty } \, \left(x-x\right)=0$$ (II) $$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$ $$\lim_{x \to \infty } \, \left(\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)*\frac{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}\right)=$$ $$\lim_{x \to \infty } \, \frac{2x}{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}=$$ $$\lim_{x \to \infty } \, \frac{2x}{\left(x\sqrt{1+1/x}+x\sqrt{1-1/x}\right)}=$$ $$\lim_{x \to \infty } \, \frac{2x}{\left(x\sqrt{1}+x\sqrt{1}\right)}=\lim_{x \to \infty } \, \frac{2x}{2x}=1$$

I found these two ways to evaluate this limit. I know the answer is 1. The first one is surely wrong. The question is: why? What is wrong there?

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Because $0\times\infty$ could be anything ... This is basically a situation where you need the second term asymptotic : Your are not allowed to take the limit for one part, and not an other one, and then to continue the computation. – Student Apr 20 '12 at 23:45
See Also:… – Aryabhata Apr 20 '12 at 23:52
up vote 7 down vote accepted

You took out the $1/x$ part. Surely $1/x\to0$ in the limit, so it may seem you can evaluate it to $0$ and then look at the rest of the function in the limit all hunky-dory, but consider applying that idea to:

$$1=\lim_{x\to\infty} \left(x\cdot\frac{1}{x}\right)=\lim_{x\to\infty}\big(x\cdot0\big) =\lim\,0=0.$$

It doesn't work!

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Ha ha, I like it :) – Student Apr 20 '12 at 23:42
@LBg: And, to get the answer, use the first two terms of the Taylor series $\sqrt{1 \pm \frac 1x} \approx 1\pm \frac 1{2x}$. The first terms cancel to avoid getting infinity and the next ones add to give what you want. The reason the second works is that the denominator avoids the cancellation. – Ross Millikan Apr 21 '12 at 3:13

By the same erroneous logic you'd conclude $\rm\:2\: =\: x(1+1/x)-x(1-1/x)\to\: 0$

In your example $\rm\:\displaystyle \sqrt{1+\frac{1}x}-\sqrt{1-\frac{1}x}\ =\ \frac{1}x + \frac{1}{8\: x^3} +\: \cdots\:$ which makes the error clear.

If the dominant terms in a sum of series cancel, then you need to look at subsequent terms.

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