Granica ciągu
: 28 maja 2010, 09:01
\(a_n= \sqrt[n]{2^n+3^{n-1}+2}\)
\(\sqrt[n]{2^n} \ \le \ \sqrt[n]{2^n+3^{n-1}+2} \ \le \ \sqrt[n]{2^n+2^n+2^n}\)
\(\lim_{n\to \infty } \sqrt[n]{2^n} = 2
\lim_{n\to \infty } \sqrt[n]{3 \cdot 2^n} = 2\)
stąd
\(\lim_{n\to \infty } \sqrt[n]{2^n+3^{n-1}+2} = 2\)