O limita tehnica cu radicali
Posted: Sat Dec 29, 2007 5:17 am
Sa se calculeze
\( \lim_{n\to\infty}\frac{\sqrt[n]{n}-1}{\sqrt{n+1}-\sqrt{n}} \).
\( \lim_{n\to\infty}\frac{\sqrt[n]{n}-1}{\sqrt{n+1}-\sqrt{n}} \).
\( \lim_{n\to\infty}\frac{\sqrt[n]{n}-1}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\to\infty}
\frac{e^{\frac{ln(n)}{n}}-1}{\frac{ln(n)}{n}} \cdot \frac{ln(n)}{n}\cdot (\sqrt{n+1}+\sqrt{n}) = \)
\( =\lim_{n\to\infty} \frac{ln(n)}{\sqrt{n}}\cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n}}=2 \lim_{n\to\infty} \frac{ln(n)}{\sqrt{n}} =^{CS} \)
\( = 2\lim_{n\to\infty} \frac{ln\left( 1+\frac{1}{n}\right)^n}{n(\sqrt{n+1}-\sqrt{n})}=2\lim_{n\to\infty} \frac{\sqrt{n+1}+\sqrt{n}}{n}=0 \)