# compute $\lim_{n\to \infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n})$ [duplicate]

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 Arkamis, Noah Snyder, Henry T. Horton, Norbert, EuYuOct 24 '12 at 20:23

Why wouldn't this be $\infty$, since left factor diverges and right factor is at least positive? – gt6989b Oct 24 '12 at 16:57
This is the same limit as here: Calculate $\displaystyle\lim_{n\to{+}\infty}{(\sqrt{n^{2}+n}-n)}$ – Martin Sleziak Oct 24 '12 at 16:59

A start: Multiply by $\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$.
then you have: $\sqrt{n}(\frac{1}{\sqrt{n+1}+\sqrt{n}})$ – Badshah Oct 24 '12 at 16:57
@Badshah: Turn this into $1\cdot\frac1{\sqrt{1+\frac1n}+\sqrt1}$. – Hagen von Eitzen Oct 24 '12 at 16:59
half done. Now "simplify" by $\sqrt n$: $\ \displaystyle\frac{1}{\frac{\sqrt{n+1}}{\sqrt n} +1}$ – Berci Oct 24 '12 at 17:00
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 – Badshah Oct 24 '12 at 17:04
@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. – André Nicolas Oct 24 '12 at 17:07