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

Question

How can I find this?

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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 $0$.