Oblicz granice
: 11 mar 2010, o 15:48
Mam pytanie czy to jest dobrze policzone? Jezeli nie to może jakieś wskazówki?
\(\displaystyle{ \lim_{x \to \infty } \frac{1}{ \sqrt{n ^{2}+3 }- \sqrt{n ^{2}-n } } = \lim_{x \to \infty } \frac{1}{ \sqrt{n ^{2}+3 }- \sqrt{n ^{2}-n } } * \frac{\sqrt{n ^{2}+3 }+ \sqrt{n ^{2}-n }}{\sqrt{n ^{2}+3 }+ \sqrt{n ^{2}-n }} =]\lim_{x \to \infty } \frac{\sqrt{n ^{2}+3 }+ \sqrt{n ^{2}-n }}{ {n ^{2}+3 }- {n ^{2}-n } } =-2}\)
Edit: oczywisice byl blad x dazy do nieskonczonosci
\(\displaystyle{ \lim_{x \to \infty } \frac{1}{ \sqrt{n ^{2}+3 }- \sqrt{n ^{2}-n } } = \lim_{x \to \infty } \frac{1}{ \sqrt{n ^{2}+3 }- \sqrt{n ^{2}-n } } * \frac{\sqrt{n ^{2}+3 }+ \sqrt{n ^{2}-n }}{\sqrt{n ^{2}+3 }+ \sqrt{n ^{2}-n }} =]\lim_{x \to \infty } \frac{\sqrt{n ^{2}+3 }+ \sqrt{n ^{2}-n }}{ {n ^{2}+3 }- {n ^{2}-n } } =-2}\)
Edit: oczywisice byl blad x dazy do nieskonczonosci