Question

Задание приложено...

Answer 1

Ответ:

Пределы:

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

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

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

Примечание:

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

Объяснение:

1)

$\displaystyle \lim_{n \to \infty} \bigg(\frac{3n + 1}{3n} \bigg)^{n} = \lim_{n \to \infty} \bigg(\frac{3n }{3n} + \frac{1}{3n} \bigg)^{n} = \lim_{n \to \infty} \bigg(1 + \frac{1}{3n} \bigg)^{n} =$

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

2)

$\displaystyle \lim_{n \to \infty} \bigg(\frac{n + 1}{n} \bigg)^{5n + 2} = \lim_{n \to \infty} \bigg(\frac{n}{n} + \frac{1}{n} \bigg)^{5n + 2} = \lim_{n \to \infty} \bigg(1 + \frac{1}{n} \bigg)^{5n + 2} =$

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

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

$\lim_{n \to \infty} \bigg(1 +0 \bigg)^{2} = \lim_{n \to \infty} 1 = 1$.

3)

$\displaystyle \lim_{n \to \infty} \bigg (1 - \dfrac{1}{n} \bigg)^{n} = \lim_{n \to \infty} \bigg (1 + \bigg( - \dfrac{1}{n} \bigg) \bigg)^{n} = \lim_{n \to \infty} \bigg (1 + \dfrac{1}{-n} \bigg)^{n} =$

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