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I need to find simple limit: $$\lim_{n\to\infty}{\sqrt{n+1}-\sqrt{n}}$$

I can't use L'Hôpital's rule, which I think is natural for such problems. How to calculate such limit using only basic methods (definition and simple transformations etc.)?

It's one problem (from 50 problems) I don't know how to solve from my homework.

Thanks for help.

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3 Answers 3

If you really mean what's on the post, you can show the limit diverges to $+\infty$. I'm pretty sure you mean $$\lim \;\sqrt{n+1}-\sqrt n$$

Then note that

$$\sqrt{n+1}-\sqrt n=\frac 1 {\sqrt{n+1}+\sqrt n}$$

The rest is pretty straightforward.

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thanks, i edited my question :). –  golert Dec 6 '12 at 23:56

Multiply by the radical conjugate:

$$\lim_{n\to\infty}\left(\frac{(\sqrt{n+1}-\sqrt{n})\cdot(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}\right)$$

$$\lim_{n\to\infty}\left(\frac{n+1-n}{(\sqrt{n+1}+\sqrt{n})}\right)$$

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I am assuming you don't really mean what's in the post, but rather what Peter Tamaroff assumed in his comment (and later, post, as I see now) –  anorton Dec 6 '12 at 23:55

Try conjugate multiplication - $\sqrt{n+1}+\sqrt n$. I assume, that you mean $\sqrt{n+1}-\sqrt n$, otherwise it's $ \infty $.

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