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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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    See Also: http://math.stackexchange.com/questions/30040/limits-how-to-evaluate-lim-limits-x-rightarrow-infty-sqrtnxna-n-12012-04-20

2 Answers 2

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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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    @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.2012-04-21
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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.