Verified answer

Ответ:

[tex]\boxed{ \boldsymbol{ \lim_{n \to \infty} \dfrac{b}{a^{n}} = 0; a > 1 } }[/tex]

Теоремы: (при условии, что [tex]a_{n},b_{n}[/tex] - сходящиеся последовательности)

Предел суммы:

[tex]\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n[/tex]

Предел произведения:

[tex]\lim_{n \to \infty} (a_n b_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n[/tex]

Предел частного:

[tex]\lim_{n \to \infty} \dfrac{a_{n}}{b_{n}} = \dfrac{\lim_{n \to \infty} a_n }{\lim_{n \to \infty} b_n }[/tex] при условии, что [tex]b_{n} \neq 0; \lim_{n \to \infty} b_{n} \neq 0[/tex]

34.5

1) [tex]\displaystyle \lim_{n \to \infty} \bigg ( \frac{2n + 1}{3 - 2n} \cdot \frac{n + 3}{4n + 5} \bigg) = \lim_{n \to \infty} \frac{2n + 1}{3 - 2n} \cdot \lim_{n \to \infty} \frac{n + 3}{4n + 5} =[/tex]

[tex]\displaystyle = \lim_{n \to \infty} \frac{\dfrac{2n + 1}{n} }{\dfrac{3 - 2n}{n} } \cdot \lim_{n \to \infty} \frac{\dfrac{n + 3}{n} }{\dfrac{4n + 5}{n} } = \lim_{n \to \infty} \frac{2 + \dfrac{1}{n} }{\dfrac{3}{n} -2 } \cdot \lim_{n \to \infty} \frac{1 + \dfrac{3}{n} }{4 + \dfrac{5}{n} } =[/tex]

[tex]\displaystyle = \frac{ \lim_{n \to \infty} \bigg ( 2 + \dfrac{1}{n} \bigg) }{\lim_{n \to \infty} \bigg( \dfrac{3}{n} -2 \bigg) } \cdot \lim_{n \to \infty} \frac{\lim_{n \to \infty} \bigg ( 1 + \dfrac{3}{n} \bigg) }{\lim_{n \to \infty} \bigg( 4 + \dfrac{5}{n} \bigg) } =[/tex]

[tex]= \displaystyle \frac{\lim_{n \to \infty} 2 + \lim_{n \to \infty} \dfrac{1}{n} }{\lim_{n \to \infty}\dfrac{3}{n} - \lim_{n \to \infty}2 } \cdot \frac{\lim_{n \to \infty}1 +\lim_{n \to \infty} \dfrac{3}{n} }{\lim_{n \to \infty}4 +\lim_{n \to \infty} \dfrac{5}{n} } = -0,25[/tex]

2)

[tex]\displaystyle \lim_{n \to \infty} \frac{2n(3n - 1)(n + 4)}{(4 + 5n)(2n + 1)(n + 1)} = \lim_{n \to \infty} \frac{2n(3n - 1)}{(4 + 5n)(2n + 1)} \cdot \lim_{n \to \infty} \frac{(n + 4)}{(n + 1)} = 0,6[/tex]

а) [tex]\displaystyle \lim_{n \to \infty} \frac{2n(3n - 1)}{(4 + 5n)(2n + 1)} = \lim_{n \to \infty} \frac{6n^{2} - 2n}{8n + 4 + 10n^{2} + 5n} = \lim_{n \to \infty} \frac{6n^{2} - 2n}{ 10n^{2} + 13n + 4} =[/tex]

[tex]\displaystyle = \lim_{n \to \infty} \frac{\dfrac{6n^{2} - 2n}{n^{2}} }{ \dfrac{10n^{2} + 13n + 4}{n^{2}} } = \lim_{n \to \infty} \frac{6 - \dfrac{2}{n} }{10 + \dfrac{13}{n} + \dfrac{4}{n^{2}} } = \frac{ \lim_{n \to \infty} \bigg( 6 - \dfrac{2}{n} \bigg) }{ \lim_{n \to \infty} \bigg(10 + \dfrac{13}{n} + \dfrac{4}{n^{2}} \bigg) } =[/tex]

[tex]\displaystyle = \frac{ \lim_{n \to \infty} 6 - \lim_{n \to \infty} \dfrac{2}{n} }{\lim_{n \to \infty}10 + \lim_{n \to \infty}\dfrac{13}{n} + \lim_{n \to \infty} \dfrac{4}{n^{2}} } = \frac{6 - 0}{10 + 0 + 0 } = \frac{6}{10} = \frac{3}{5} = 0,6[/tex]

б) [tex]\displaystyle \lim_{n \to \infty} \frac{(n + 4)}{(n + 1)} = \lim_{n \to \infty} \frac{\dfrac{n + 4}{n} }{\dfrac{n + 1}{n} } = \lim_{n \to \infty} \frac{1 + \dfrac{4}{n} }{1 + \dfrac{1}{n} } = \frac{ \lim_{n \to \infty} \bigg( 1 + \dfrac{4}{n} \bigg) }{\lim_{n \to \infty} \bigg( 1 + \dfrac{1}{n} \bigg)} =[/tex]

[tex]\displaystyle = \frac{ \lim_{n \to \infty} 1 + \lim_{n \to \infty} \dfrac{4}{n} }{ \lim_{n \to \infty} 1 + \lim_{n \to \infty}\dfrac{1}{n} } = 1[/tex]

3) [tex]\displaystyle \lim_{n \to \infty} \frac{(n + 1)(n - 2)(2n + 3)}{(4n - 1)(n + 3)(5n - 2)} = \lim_{n \to \infty} \frac{n + 1}{4n - 1} \cdot \lim_{n \to \infty} \frac{n - 2}{n + 3} \cdot \lim_{n \to \infty} \frac{2n + 3}{5n - 2} = 0,1[/tex]

а) [tex]\displaystyle \lim_{n \to \infty} \frac{n + 1}{4n - 1} = \lim_{n \to \infty} \frac{\dfrac{n + 1}{n} }{\dfrac{4n - 1}{n} } = \lim_{n \to \infty} \frac{1 + \dfrac{1}{n} }{4 - \dfrac{1}{n} } = \frac{\lim_{n \to \infty} \bigg( 1 + \dfrac{1}{n} \bigg) }{\lim_{n \to \infty} \bigg( 4 - \dfrac{1}{n} \bigg) } =[/tex]

[tex]\displaystyle = \frac{ \lim_{n \to \infty} 1 + \lim_{n \to \infty} \dfrac{1}{n} }{\lim_{n \to \infty} 4 - \lim_{n \to \infty} \dfrac{1}{n} } = \frac{1}{4}[/tex]

б) [tex]\displaystyle \lim_{n \to \infty} \frac{n - 2}{n + 3} = \lim_{n \to \infty} \frac{\dfrac{n - 2}{n} }{\dfrac{n + 3}{n} } = \lim_{n \to \infty} \frac{1 - \dfrac{2}{n} }{1 + \dfrac{3}{n} } = \frac{\lim_{n \to \infty} \bigg( 1 - \dfrac{2}{n} \bigg) }{\lim_{n \to \infty} \bigg( 1 + \dfrac{3}{n} \bigg) } =[/tex]

[tex]\displaystyle = \frac{ \lim_{n \to \infty} 1 - \lim_{n \to \infty} \dfrac{2}{n} }{\lim_{n \to \infty} 1 + \lim_{n \to \infty} \dfrac{3}{n} } = 1[/tex]

в) [tex]\displaystyle \lim_{n \to \infty} \frac{2n + 3}{5n - 2} = \lim_{n \to \infty} \frac{\dfrac{2n + 3}{n} }{\dfrac{5n - 2}{n} } = \lim_{n \to \infty} \frac{2 + \dfrac{3}{n} }{5 - \dfrac{2}{n} } = \frac{ \lim_{n \to \infty}\bigg ( 2 + \dfrac{3}{n} \bigg)}{ \lim_{n \to \infty}\bigg (5 - \dfrac{2}{n} \bigg)} =[/tex]

[tex]\displaystyle = \frac{ \lim_{n \to \infty} 2 + \lim_{n \to \infty} \dfrac{3}{n} }{ \lim_{n \to \infty} 5 - \lim_{n \to \infty} \dfrac{2}{n} } = \frac{2}{5}[/tex]