# 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
This nice hint maskes calculus much simpler!+1 صباح الخير أخي قل لي هل تجيد العربية؟ –  Sami Ben Romdhane Dec 8 '13 at 7:03
@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

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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