Sa se calculeze
\( \lim_{n\to\infty}\frac{\sqrt[n]{n}-1}{\sqrt{n+1}-\sqrt{n}} \).
O limita tehnica cu radicali
Moderators: Bogdan Posa, Laurian Filip, Beniamin Bogosel, Radu Titiu, Marius Dragoi
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
O limita tehnica cu radicali
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
-
Radu Titiu
- Thales
- Posts: 155
- Joined: Fri Sep 28, 2007 5:05 pm
- Location: Mures \Bucuresti
\( \lim_{n\to\infty}\frac{\sqrt[n]{n}-1}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\to\infty}
\frac{e^{\frac{ln(n)}{n}}-1}{\frac{ln(n)}{n}} \cdot \frac{ln(n)}{n}\cdot (\sqrt{n+1}+\sqrt{n}) = \)
\( =\lim_{n\to\infty} \frac{ln(n)}{\sqrt{n}}\cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n}}=2 \lim_{n\to\infty} \frac{ln(n)}{\sqrt{n}} =^{CS} \)
\( = 2\lim_{n\to\infty} \frac{ln\left( 1+\frac{1}{n}\right)^n}{n(\sqrt{n+1}-\sqrt{n})}=2\lim_{n\to\infty} \frac{\sqrt{n+1}+\sqrt{n}}{n}=0 \)
\frac{e^{\frac{ln(n)}{n}}-1}{\frac{ln(n)}{n}} \cdot \frac{ln(n)}{n}\cdot (\sqrt{n+1}+\sqrt{n}) = \)
\( =\lim_{n\to\infty} \frac{ln(n)}{\sqrt{n}}\cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n}}=2 \lim_{n\to\infty} \frac{ln(n)}{\sqrt{n}} =^{CS} \)
\( = 2\lim_{n\to\infty} \frac{ln\left( 1+\frac{1}{n}\right)^n}{n(\sqrt{n+1}-\sqrt{n})}=2\lim_{n\to\infty} \frac{\sqrt{n+1}+\sqrt{n}}{n}=0 \)
A mathematician is a machine for turning coffee into theorems.