# Find the limit: $\displaystyle\lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x}$.

Find the following limit: $$\lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x}$$

I tried to simplify using conjugation. This gave me the following: $$\lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}$$

When I plug in the $\infty$, I'm left with $\frac{\infty}{\infty}$. Did I mess up somewhere, or does the limit not exist?

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Since you have $\infty/\infty,$ looks like a job for $l'Hopital.$ Have you tried it? – Igor Rivin Dec 7 '13 at 19:13
I'm aware of the rule, but this is part of review for earlier units in my Calculus class. I don't think my professor wants us to use that on this particular question. – Ahounsel Dec 7 '13 at 19:21
– lab bhattacharjee Dec 8 '13 at 4:10

You are almost there:

$$\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} = \frac{2x}{x \sqrt{1+\frac{1}{x}}+x\sqrt{1-\frac{1}{x}}}$$

so...

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Hint:

$$ax^2+bx+c\approx ax^2, ~~x\to\infty$$ so if $x\to+\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=x$ and if $x\to-\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=-x$

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I always like this intuitive approach! +1 – amWhy Dec 7 '13 at 22:47
@SamiBenRomdhane: I know some words up to the sharia. Not much. ;-) – Babak S. Dec 8 '13 at 8:03
@B.S. Of course you can ask. And I'll try to help in any way I can! – amWhy Dec 8 '13 at 13:22
What was the question you asked? Is he still angry with you about the question? Or are you afraid to bring it up again because of his anger when you asked the question? Did he give you an answer? – amWhy Dec 8 '13 at 13:39
I can't understand why he got angry! So you asked a question here on MSE and he is upset because of that? That is absurd? It is a very general question, and hardly gives away anything in the way of any proposal you are working on. I find it hard to understand why your supervisor is upset. So are you wondering how/if you can (if you can) delete the question? I don't think that's possible. – amWhy Dec 8 '13 at 13:57

For $x\ge 1$ we have $$\sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}.$$ It follows that $$\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1.$$

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