limit as x-> infinity

**mistykz** · August 31st 2007, 01:37 PM

okay, so the question is what is the limit as x-> infinity of (sqrt(x^2+x))-(sqrt(x^2-x))?
I know it is 1, but how do i show work / come to that conclusion?

**Krizalid** · August 31st 2007, 01:40 PM

What about if you multiply the limit by $\frac{\sqrt {x^2 + x} + \sqrt {x^2 - x}}{\sqrt {x^2 + x} + \sqrt {x^2 - x}}$ ?

**TKHunny** · August 31st 2007, 01:42 PM

Originally Posted by mistykz

okay, so the question is what is the limit as x-> infinity of (sqrt(x^2+x))-(sqrt(x^2-x))?
I know it is 1, but how do i show work / come to that conclusion?

A nice trick, probably contrary to what has been beaten into your brain since Algebra I. UNrationalize it!!

In other words, multiply by $\frac{\sqrt{x^{2}+x}+\sqrt{x^{2}-x}}{\sqrt{x^{2}+x}+\sqrt{x^{2}-x}}$ .

Resolve the numerator and wonderful things should appear. At least there should be somewhere to go with it.

**red_dog** · August 31st 2007, 01:45 PM

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+x}+\sqrt{ x^2-x}}=$
$\displaystyle =\lim_{x\to\infty}\frac{2x}{x\left(\sqrt{1+\frac{1 }{x}}+\sqrt{1-\frac{1}{x}}\right)}=\lim_{x\to\infty}\frac{2}{\sq rt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1$

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