Wiedząc, że...- Zadanie 7.149: Matematyka 2. Poziom rozszerzony. Po gimnazjum

Rozwiązanie

Własności granic ciągów zbieżnych:

$n \to \infty lim (a_{n} + b_{n}) = n \to \infty lim a_{n} + n \to \infty lim b_{n}$

$n \to \infty lim (a_{n} - b_{n}) = n \to \infty lim a_{n} - n \to \infty lim b_{n}$

$n \to \infty lim (a_{n} \cdot b_{n}) = n \to \infty lim a_{n} \cdot n \to \infty lim b_{n}$

$n \to \infty lim (\frac{a _{n}}{b _{n}}) = \frac{n \to \infty lim a _{n}}{n \to \infty lim b _{n}}, n \to \infty lim b_{n} \neq = 0$

$n \to \infty lim (p \cdot a_{n}) = p \cdot n \to \infty lim a_{n}$

$n \to \infty lim p = p$

$a) n \to \infty lim (a_{n} + 3 b_{n}) = n \to \infty lim a_{n} + n \to \infty lim 3 b_{n} = n \to \infty lim a_{n} + 3 \cdot n \to \infty lim b_{n} = 2 + 3 \cdot \frac{2}{3} = 2 + 2 = 4$

$b) n \to \infty lim \frac{a _{n}}{b _{n}} = \frac{n \to \infty lim a _{n}}{n \to \infty lim b _{n}} = \frac{2}{\frac{2}{3}} = 2 \cdot \frac{3}{2} = 3$

$c) n \to \infty lim a_{n} \cdot b_{n} = n \to \infty lim (a_{n} \cdot b_{n}) = n \to \infty lim a_{n} \cdot n \to \infty lim b_{n} = 2 \cdot \frac{2}{3} = 2 \cdot \frac{2}{3} = \frac{2}{3} \cdot \frac{3}{3} = \frac{2 3}{3}$

$d) n \to \infty lim (a_{n} + \frac{1}{b _{n}}) = n \to \infty lim a_{n} + n \to \infty lim \frac{1}{b _{n}} = n \to \infty lim a_{n} + \frac{n \to \infty lim 1}{n \to \infty lim b _{n}} = 2 + \frac{1}{\frac{2}{3}} = 2 + \frac{3}{2} = \frac{4}{2} + \frac{3}{2} = \frac{7}{2} = 3 \frac{1}{2}$

$e) n \to \infty lim (4 a_{n} - 6 b_{n}) = n \to \infty lim 4 a_{n} - n \to \infty lim 6 b_{n} = 4 \cdot n \to \infty lim a_{n} - 6 \cdot n \to \infty lim b_{n} = 4 \cdot 2 - 6 \cdot \frac{2}{3} = 8 - 2 \cdot 2 = 8 - 4 = 4$

$f) n \to \infty lim \frac{2 \cdot a _{n} \cdot b _{n}}{a _{n} + b _{n}} = \frac{n \to \infty lim ( 2 \cdot a _{n} \cdot b _{n} )}{n \to \infty lim ( a _{n} + b _{n} )} = \frac{2 \cdot n \to \infty lim ( a _{n} \cdot b _{n} )}{n \to \infty lim a _{n} + n \to \infty lim b _{n}} = \frac{2 \cdot n \to \infty lim a _{n} \cdot n \to \infty lim b _{n}}{n \to \infty lim a _{n} + n \to \infty lim b _{n}} = \frac{2 \cdot 2 \cdot \frac{2}{3}}{2 + \frac{2}{3}} = \frac{\frac{8}{3}}{\frac{8}{3}} = 1$

$g) n \to \infty lim (\frac{a _{n} + b _{n}}{2}) = \frac{n \to \infty lim ( a _{n} + b _{n} )}{n \to \infty lim 2} = \frac{n \to \infty lim a _{n} + n \to \infty lim b _{n}}{2} = \frac{2 + \frac{2}{3}}{2} = \frac{2 \cdot ( 1 + \frac{1}{3} )}{2} = 1 + \frac{1}{3} = \frac{4}{3}$

$h) n \to \infty lim (a_{n}^{2} + b_{n}^{2}) = n \to \infty lim a_{n}^{2} + n \to \infty lim b_{n}^{2} = (n \to \infty lim a_{n})^{2} + (n \to \infty lim b_{n})^{2} = 2^{2} + (\frac{2}{3})^{2} = 4 + \frac{4}{9} = \frac{36}{9} + \frac{4}{9} = \frac{40}{9} = 4 \frac{4}{9}$

Ernest Jamka

Nauczyciel matematyki

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