# how to find $\lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$ [closed]

How can I find this?

$\lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$

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## closed as off-topic by TMM, Lost1, Davide Giraudo, Old John, mathematics2x2lifeJan 25 '14 at 20:33

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$\lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right) =\lim\limits_{x \to \infty} \left(\sqrt{x^2 +1}-x +\sqrt{4x^2 + 1}-2x - (\sqrt{9x^2 + 1}-3x)\right) =$ $\lim\limits_{x \to \infty}\frac{1}{\sqrt{x^2 +1}+x}+\lim\limits_{x \to \infty}\frac{1}{\sqrt{4x^2 + 1}+2x} -\lim\limits_{x \to \infty}\frac{1}{\sqrt{9x^2 + 1}+3x} = 0 + 0 + 0 = 0$

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What is unclear? – medicu Jan 25 '14 at 19:25
Probably because you forgot the minus sign but otherwise I find your answer very good. I just picked ncmathaddict's answer because he answered first and it just didn't occur to me to add the extra $x_s$ – Veritas Jan 25 '14 at 19:42
I threw in the minus sign and a few limits. You can rollback if it's not satisfactory. – hardmath Jan 25 '14 at 21:40
It wasn't me who downvoted. – Veritas Jan 25 '14 at 22:03

Here is another tack. If $a > 0$, $${\sqrt{a^2 x^2 + 1} - ax } = {1\over{\sqrt{a^2 x^2 + 1} + ax }}= O\left ({1\over x}\right).$$

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Very nice this trick! – MathOverview Jan 25 '14 at 19:22
This did it! It didn't occur to me to add and subtract the extra $x_s$. – Veritas Jan 25 '14 at 19:30

Since for any $A>0$ $$\sqrt{A^2 x^2+1}-A|x| = \frac{1}{A|x|+\sqrt{A^2 x^2+1}}<\frac{1}{2A|x|}$$ holds, we have: $$\left|\sqrt{x^2+1}+\sqrt{4x^2+1}-\sqrt{9x^2+1}\right|=\left|\sqrt{x^2+1}-|x|+\sqrt{4x^2+1}-2|x|-\sqrt{9x^2+1}+3|x|\right|\leq \left|\sqrt{x^2+1}-|x|\right|+\left|\sqrt{4x^2+1}-2|x|\right|+\left|\sqrt{9x^2+1}-3|x|\right|<\frac{1}{|x|}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right),$$ hence the limit is $0$.

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A heuristic result:

Note that $x$ ~$\sqrt{x^2 + 1}$ as $x \to \infty$, and similar results hold for the other terms. Thus the limit is seen to be zero.

To make the result more rigorous, observe what happens when you expand the result in Elias' answer using the binomial series.

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