granice ciagów
: 4 paź 2007, o 21:35
Znaleśc granice ciągów:
\(\displaystyle{ \lim_{n\to } ft( \sqrt{(n^{2}+1)-n}\right)
\\ \lim_{n\to } ft( \sqrt[3]{n^{3}+2n^{2}}-n\right)
\\ \lim_{n\to } ft(\frac{ \sqrt{n+1} - \sqrt{n}}{ \sqrt{2n+1} - \sqrt{n+1}}\right)
\\ \lim_{n\to } ft(\frac{n}{2^{n}}\right)
\\ \lim_{n\to } ft(\frac{2+n}{3+n}\right)^{n}
\\ \lim_{n\to } ft(\frac{2n+1}{3n+1}\right)^{n}
\\ \lim_{n\to } ft \frac{2^{\frac{1}{n}}-1}{2^{\frac{1}{n}}+1}\right
\\ \lim_{n\to } ft(n(ln(n+1)-ln(n))\right)}\)
\(\displaystyle{ \lim_{n\to } ft( \sqrt{(n^{2}+1)-n}\right)
\\ \lim_{n\to } ft( \sqrt[3]{n^{3}+2n^{2}}-n\right)
\\ \lim_{n\to } ft(\frac{ \sqrt{n+1} - \sqrt{n}}{ \sqrt{2n+1} - \sqrt{n+1}}\right)
\\ \lim_{n\to } ft(\frac{n}{2^{n}}\right)
\\ \lim_{n\to } ft(\frac{2+n}{3+n}\right)^{n}
\\ \lim_{n\to } ft(\frac{2n+1}{3n+1}\right)^{n}
\\ \lim_{n\to } ft \frac{2^{\frac{1}{n}}-1}{2^{\frac{1}{n}}+1}\right
\\ \lim_{n\to } ft(n(ln(n+1)-ln(n))\right)}\)