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

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4 Answers 4

up vote 2 down vote accepted

$ \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 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 at 19:42
I threw in the minus sign and a few limits. You can rollback if it's not satisfactory. –  hardmath Jan 25 at 21:40
It wasn't me who downvoted. –  Veritas Jan 25 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! –  Elias Jan 25 at 19:22
This did it! It didn't occur to me to add and subtract the extra $x_s$. –  Veritas Jan 25 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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