Zadatak 0838
- Detalji
- Kategorija: Granična vrednost funkcije
- Objavljeno 27 novembar 2013
- Autor Super User
- Pogodaka: 976
Test: MKsIVBGI_073
Izračunaj graničnu vrednost niza: $$\lim \limits_{n \to +\infty }\left ( \sqrt{n^2+2n+2}-\sqrt{n^2-4n+3} \right )$$
Rešenje:
$$...=\lim \limits_{n \to +\infty }\left ( \sqrt{n^2+2n+2}-\sqrt{n^2-4n+3} \right )\cdot \frac{\left ( \sqrt{n^2+2n+2}+\sqrt{n^2-4n+3} \right )}{\left ( \sqrt{n^2+2n+2}+\sqrt{n^2-4n+3} \right )}$$
$$=\lim \limits_{n \to +\infty }\frac{n^2+2n+2-n^2+4n-3}{\left ( \sqrt{n^2+2n+2}+\sqrt{n^2-4n+3} \right )}$$
$$=\lim \limits_{n \to +\infty }\frac{n(6-\frac{1}{n})}{n(\sqrt{1+\frac{2}{n}+\frac{2}{n^2}}+\sqrt{1-\frac{4}{n}+\frac{3}{n^2}})}$$
$$=\frac{6}{2}=3$$