Naci granicnu vrednost...

$e^{\lim_{n\to \infty } n\frac{n-\sqrt{n^2+ 2n}}{n+\sqrt{n^2+ 2n}}}=e^{\lim_{n\to \infty } n\frac{n-\sqrt{n^2+ 2n}}{n+\sqrt{n^2+ 2n}}\frac{n+\sqrt{n^2+ 2n}}{n+\sqrt{n^2+ 2n}}}=e^{\lim_{n\to \infty } n\frac{-2n}{\left(n+\sqrt{n^2+ 2n}\right)^2}}=e^{\lim_{n\to \infty } \frac{-2n^2}{n^2+2n\sqrt{n^2+ 2n}+n^2+2n}}=$

$=e^{\lim_{n\to \infty } \frac{-2n^2}{n^2\left(1+\frac{2}{n}\sqrt{n^2+ 2n}+1+\frac{2}{n}\right)}}=e^{\lim_{n\to \infty } \frac{-2}{\left(1+2\sqrt{1+ \frac{2}{n}}+1+\frac{2}{n}\right)}}=e^{\frac{-2}{4}}=e^{-1/2}=\frac{1}{\sqrt{e}}$

Edit: Ako se ne vidi evo ovako:

$n\frac{n-\sqrt{n^2+ 2n}}{n+\sqrt{n^2+ 2n}}=n\frac{n-\sqrt{n^2+ 2n}}{n+\sqrt{n^2+ 2n}}\frac{n+\sqrt{n^2+ 2n}}{n+\sqrt{n^2+ 2n}}=n\frac{-2n}{\left(n+\sqrt{n^2+ 2n}\right)^2}=\frac{-2n^2}{n^2+2n\sqrt{n^2+ 2n}+n^2+2n}=\frac{-2n^2}{n^2\left(1+\frac{2}{n}\sqrt{n^2+ 2n}+1+\frac{2}{n}\right)}=\frac{-2}{\left(1+2\sqrt{1+ \frac{2}{n}}+1+\frac{2}{n}\right)}$

Potpis

A mathematician is a blind man in a dark room looking for a black cat which isn't there.

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