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Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
Calculate $\displaystyle\lim_{n\to{+}\infty}{(\sqrt{n^{2}+n}-n)}$

how should I approach the following limit? $$\lim_{n\to \infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n})$$

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marked as duplicate by Emily, Noah Snyder, Henry T. Horton, Norbert, EuYu Oct 24 '12 at 20:23

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A start: Multiply by $\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$.

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  • $\begingroup$ then you have: $\sqrt{n}(\frac{1}{\sqrt{n+1}+\sqrt{n}})$ $\endgroup$ – Badshah Oct 24 '12 at 16:57
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    $\begingroup$ @Badshah: Turn this into $1\cdot\frac1{\sqrt{1+\frac1n}+\sqrt1}$. $\endgroup$ – Hagen von Eitzen Oct 24 '12 at 16:59
  • $\begingroup$ half done. Now "simplify" by $\sqrt n$: $\ \displaystyle\frac{1}{\frac{\sqrt{n+1}}{\sqrt n} +1}$ $\endgroup$ – Berci Oct 24 '12 at 17:00
  • $\begingroup$ oke I see, you multiplied the numerator and denominator with $1/\sqrt{n}$. then the limit becomes 1/2. Is it possible to say that the limit of $\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$ equals, say, z and take the inverse, so $\frac{1}{z}=\lim_{n\to\infty}\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n}}$? then it follows that z=1/2 $\endgroup$ – Badshah Oct 24 '12 at 17:04
  • $\begingroup$ @Badshah: This sort of move can be dangerous, because it assumes that the limit exists. But if for some reason you know that the limit exists and is not $0$, it is fine. $\endgroup$ – André Nicolas Oct 24 '12 at 17:07

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