Oblicz granice.
a) [tex]\lim_{n \to \infty} \sqrt[n]{7+sin n}[/tex]
b) [tex]a_{n} = \frac{2^{n} + 5^{n} }{5^{n}-2^{n} }[/tex]
a) [tex]\lim_{n \to \infty} \sqrt[n]{7+sin n}[/tex]
b) [tex]a_{n} = \frac{2^{n} + 5^{n} }{5^{n}-2^{n} }[/tex]
Odpowiedź:
a ) [tex]\lim_{n \to \infty} \sqrt[n]{7 + sin n} = 1[/tex]
b ) [tex]a_n = \frac{2^n + 5^n}{5^n- 2^n} = \frac{5^n + 2^n}{5^n- 2^n}[/tex] [tex]= \frac{1 + (\frac{2}{5})^n}{1 - (\frac{2}{5})^n}[/tex]
więc [tex]\lim_{n \to \infty} a_n = \frac{1 + 0}{1 - 0} = 1[/tex]
Szczegółowe wyjaśnienie: