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How can I show/explain the following limit?

$$\lim_{x\to\infty} \;x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$$

Some trivial transformation seems to be eluding me.

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hint: multiply by the conjugate $\sqrt{}+\sqrt{}$. – Raymond Manzoni Apr 1 '13 at 11:46
Every time you use double dollar symbol on titles, a unicorn dies! – Pedro Tamaroff Apr 1 '13 at 11:50
@PeterTamaroff Someone had to tell you that unicorns don't exist. Sorry. – 1015 Apr 1 '13 at 11:59
@julien: Nullius in verba, prove it! :-) – Raymond Manzoni Apr 1 '13 at 12:02
up vote 8 down vote accepted

The expression can be multiplied with its conjugate and then:

$$\begin{align} \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right) &= \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)\left(\frac{\sqrt{x^2-1}+\sqrt{x^2+1}}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} x\left(\frac{x^2-1-x^2-1}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} x\left(\frac{-2}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} \frac{-2}{\frac{\sqrt{x^2-1}}{x} + \frac{\sqrt{x^2+1}}{x}} \cr &=\lim_{x\to\infty} \frac{-2}{\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}} \cr &=\lim_{x\to\infty} \frac{-2}{\sqrt{1-0} + \sqrt{1-0}} \cr &=\lim_{x\to\infty} \frac{-2}{1+1} \cr &= -1\end{align}$$

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Putting $\frac1{x^2}=h$

So, $h\to0$ as $x\to\infty$



$$\text{Now, }\sqrt{1-h^2}-\sqrt{1+h^2}=\frac{(1-h^2)-(1+h^2)}{\sqrt{1-h^2}+\sqrt{1+h^2}}=\frac{-2h^2}{\sqrt{1-h^2}+\sqrt{1+h^2}}$$

$$\implies \frac{\sqrt{1-h}-\sqrt{1+h}}h=\frac{-2}{\sqrt{1-h}+\sqrt{1+h}}$$




$$=-\lim_{h\to0}\frac{\left(1+\frac h2+O(h^2)\right)-\left(1-\frac h2+O(h^2)\right)}{h^2}$$


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I don't understand the trick with 1/n = h and h->0 as x->inf. Can you please explain it more in detail? (I also guess that it was meant to be 1/x = h?) – A123321 Apr 1 '13 at 12:04
"Putting $\frac {1}{n}=h$" should be replaced by "Putting $\frac {1}{x}=h$". – learner Apr 1 '13 at 12:26
@learner, thanks for your observation. – lab bhattacharjee Apr 1 '13 at 14:10

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