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Ответ:

Пределы:

1) [tex]\boxed{ \boldsymbol { \displaystyle \lim_{n \to \infty} \frac{n^{2} + 3n + 15}{2n^{2} - n + 100} = 0,5 } }[/tex]

2) [tex]\boxed { \boldsymbol { \displaystyle \lim_{n \to \infty} \frac{n - 3n^{2} + 44}{n^{2} + 5n -7} = -3 } }[/tex]

3) [tex]\boxed { \boldsymbol { \displaystyle \lim_{n \to \infty} \frac{4n^{5} + 5n^{4} + 3n - 2}{9n^{5} + n^{3} - 1} = \frac{4}{9} } }[/tex]

Примечание:

[tex]\boxed{ \boldsymbol{ \lim_{n \to \infty} \dfrac{b}{a^{n}} = 0 ; a > 0, b \in \mathbb R} }[/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].

Объяснение:

34.3

1)

[tex]\displaystyle \lim_{n \to \infty} \frac{n^{2} + 3n + 15}{2n^{2} - n + 100} = \lim_{n \to \infty} \frac{\cfrac{n^{2} + 3n + 15}{n^{2}} }{\cfrac{2n^{2} - n + 100}{n^{2}} } = \cfrac{ \lim_{n \to \infty} \bigg ( 1 + \dfrac{3}{n} + \dfrac{15}{n^{2}} \bigg)}{\lim_{n \to \infty} \bigg ( 2 - \dfrac{1}{n} + \dfrac{100}{n^{2}} \bigg)} =[/tex]

[tex]= \cfrac{ \lim_{n \to \infty} 1 + \lim_{n \to \infty}\dfrac{3}{n} +\lim_{n \to \infty} \dfrac{15}{n^{2}} }{ \lim_{n \to \infty} 2 - \lim_{n \to \infty} \dfrac{1}{n} +\lim_{n \to \infty} \dfrac{100}{n^{2}}} = \dfrac{1 + 0 + 0}{2 -0 + 0} = \dfrac{1}{2} =0,5[/tex]

2)

[tex]\displaystyle \lim_{n \to \infty} \frac{n - 3n^{2} + 44}{n^{2} + 5n -7} = \lim_{n \to \infty} \frac{\cfrac{n - 3n^{2} + 44}{n^{2}} }{\cfrac{n^{2} + 5n -7}{n^{2}} } = \cfrac{ \lim_{n \to \infty} \bigg (\dfrac{1}{n} - 3 + \dfrac{44}{n^{2}} \bigg) }{\lim_{n \to \infty} \bigg (1+ \dfrac{5}{n} - \dfrac{7}{n^{2}} \bigg)} =[/tex]

[tex]= \cfrac{ \lim_{n \to \infty} -3 + \lim_{n \to \infty}\dfrac{1}{n} +\lim_{n \to \infty} \dfrac{44}{n^{2}} }{ \lim_{n \to \infty} 1 + \lim_{n \to \infty} \dfrac{5}{n} - \lim_{n \to \infty} \dfrac{7}{n^{2}}} = \dfrac{-3 + 0 + 0}{1 + 0 - 0} = \dfrac{-3}{1} = -3[/tex]

3)

[tex]\displaystyle \lim_{n \to \infty} \frac{4n^{5} + 5n^{4} + 3n - 2}{9n^{5} + n^{3} - 1} = \lim_{n \to \infty} \cfrac{\cfrac{4n^{5} + 5n^{4} + 3n - 2}{n^{5}} }{\cfrac{9n^{5} + n^{3} - 1}{n^{5}} } =[/tex]

[tex]\displaystyle = \cfrac{ \lim_{n \to \infty} \bigg (4 + \dfrac{5}{n} + \dfrac{3}{n^{4} } - \dfrac{2}{n^{5}} \bigg)}{\lim_{n \to \infty} \bigg ( 9 +\dfrac{1}{n^{2}} - \dfrac{1}{n^{5}} \bigg)} =[/tex]

[tex]\displaystyle = \cfrac{ \lim_{n \to \infty} 4 +\lim_{n \to \infty} \dfrac{5}{n} + \lim_{n \to \infty} \dfrac{3}{n^{4} } - \lim_{n \to \infty} \dfrac{2}{n^{5}} }{\lim_{n \to \infty} 9 + \lim_{n \to \infty} \dfrac{1}{n^{2}} - \lim_{n \to \infty} \dfrac{1}{n^{5}} } = \frac{4 + 0 + 0 - 0}{9 + 0 - 0} = \frac{4}{9}[/tex].