od Daniel » Subota, 01. Decembar 2012, 12:05
Vrlo jednostavno, ne treba nikakvo racionalisanje, nego samo podeliš i brojilac i imenilac sa [inlmath]\sqrt{n}[/inlmath]:
[dispmath]\lim_{n\to\infty}\frac{\frac{\sqrt{n+1}+\sqrt{n+2}+\cdots+\sqrt{n+100}}{\sqrt n}}{\frac{\sqrt n+\sqrt{n+1}+\cdots+\sqrt{n+10}}{\sqrt n}}=[/dispmath][dispmath]=\lim_{n\to\infty}\frac{\frac{\sqrt{n+1}}{\sqrt n}+\frac{\sqrt{n+2}}{\sqrt n}+\cdots+\frac{\sqrt{n+100}}{\sqrt n}}{\frac{\sqrt n}{\sqrt n}+\frac{\sqrt{n+1}}{\sqrt n}+\cdots+\frac{\sqrt{n+10}}{\sqrt n}}=[/dispmath][dispmath]=\lim_{n\to\infty}\frac{\sqrt{\frac{n+1}{n}}+\sqrt{\frac{n+2}{n}}+\cdots+\sqrt{\frac{n+100}{n}}}{\sqrt{\frac{n}{n}}+\sqrt{\frac{n+1}{n}}+\cdots+\sqrt{\frac{n+10}{n}}}=[/dispmath][dispmath]=\lim_{n\to\infty}\frac{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+\cdots+\sqrt{1+\frac{100}{n}}}{\sqrt 1+\sqrt{1+\frac{1}{n}}+\cdots+\sqrt{1+\frac{10}{n}}}=[/dispmath]
Kada [inlmath]n\to\infty[/inlmath] tada konačan broj podeljen sa [inlmath]n[/inlmath] teži nuli, pa pišemo:
[dispmath]=\frac{\overbrace{1+1+\cdots+1}^{100}}{\underbrace{1+1+\cdots+1}_{11}}=[/dispmath][dispmath]=\frac{100}{11}[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain