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Ответ:

Пределы:

1) [tex]\boxed{ \boldsymbol{ \displaystyle \lim_{n \to \infty} \bigg(\frac{3n + 1}{3n} \bigg)^{n} = \sqrt[3]{e} } }[/tex]

2) [tex]\boxed{ \boldsymbol{ \displaystyle \lim_{n \to \infty} \bigg(\frac{n + 1}{n} \bigg)^{5n + 2} = e^{5} } }[/tex]

3) [tex]\boxed{ \boldsymbol{ \displaystyle \lim_{n \to \infty} \bigg (1 - \dfrac{1}{n} \bigg)^{n} =\frac{1}{e} } }[/tex]

Примечание:

[tex]\boxed{ \boldsymbol{ \lim_{n \to \infty} \bigg (1 + \dfrac{1}{n} \bigg)^{n} = e} }[/tex] - второй замечательный предел

Объяснение:

1)

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

[tex]\displaystyle = \lim_{n \to \infty} \sqrt[3]{\bigg(1 + \frac{1}{3n} \bigg)^{3n}} = \sqrt[3]{e}[/tex]

2)

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

[tex]\displaystyle = \lim_{n \to \infty} \bigg(1 + \frac{1}{n} \bigg)^{5n } \cdot \lim_{n \to \infty} \bigg(1 + \frac{1}{n} \bigg)^{2} = \lim_{n \to \infty} \bigg( \bigg(1 + \frac{1}{n} \bigg)^{n} \bigg)^{5}\cdot1 = e^{5}[/tex]

Рассмотрим предел [tex]\lim_{n \to \infty} \bigg(1 + \dfrac{1}{n} \bigg)^{2}[/tex], так как [tex]n \to \infty \Longrightarrow \dfrac{1}{n} \to 0[/tex], тогда

[tex]\lim_{n \to \infty} \bigg(1 +0 \bigg)^{2} = \lim_{n \to \infty} 1 = 1[/tex].

3)

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

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