Nepero numero 1

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$b_n:\mathbb{N}\setminus \{0\}\rightarrow \mathbb{R}$

la successione definita da

$b_n=\left(1+\dfrac{1}{n} \right)^n.$

Si definisce numero di Nepero

$\begin{equation*} e=\lim_{n \rightarrow +\infty}b_n \end{equation*}$

con $2<e<3$ .
In generale vale quanto segue: data una successione ${c_n}$ infinita per $n \rightarrow +\infty$ , vale che

(1) $\begin{equation*} e=\lim_{n \rightarrow +\infty }\left(1+\dfrac{1}{c_n} \right)^{c_n} \end{equation*}$

Esercizio 1. $(\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar)$ Calcolare il seguente limite, applicando (1):

(2) $\begin{equation*} \lim_{n \to +\infty} \left( 1+\dfrac{2}{n}\right) ^n. \end{equation*}$

Svolgimento esercizio 1. Risoviamo (2) applicando (1):

$\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{2}{n}\right) ^n & = \lim_{n \to +\infty} \left[ \left( 1+\dfrac{1}{\frac{n}{2}}\right) ^{\frac{n}{2}}\right]^2 = e^2. \end{aligned}$

Si conclude che:

$\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1+\dfrac{2}{n}\right) ^n=e^2.}$

Esercizio 2. $(\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar)$ Calcolare il seguente limite, applicando (1):

(3) $\begin{equation*} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^n. \end{equation*}$

Svolgimento esercizio 2. Risolviamo (3) applicando (1):

$\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^n & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{n^2}\right) ^{n^2}\right) ^{\frac{1}{n}}=e^0=1. \end{aligned}$

Si conclude che:

$\boxcolorato{analisi}{\lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^n =1.}$

Esercizio 3. $(\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar)$ Calcolare il seguente limite, applicando (1):

(4) $\begin{equation*} \lim_{n \to +\infty} \left( 1-\dfrac{3}{n}\right) ^n. \end{equation*}$

Svolgimento esercizio 3. Risolviamo (4) applicando (1):

$\begin{aligned} \lim_{n \to +\infty} \left( 1-\dfrac{3}{n}\right) ^n & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{-\dfrac{n}{3}}\right) ^{-\dfrac{n}{3}}\right) ^{-3}=e^{-3}=\dfrac{1}{e^3}. \end{aligned}$

Si conclude che:

$\boxcolorato{analisi}{\lim_{n \to +\infty} \left( 1-\dfrac{3}{n}\right) ^n=\dfrac{1}{e^3}.}$

Esercizio 4. $(\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar)$ Calcolare il seguente limite, applicando (1):

(5) $\begin{equation*} \lim_{n \to +\infty} \left( 1-\dfrac{1}{n^2}\right) ^n. \end{equation*}$

Svolgimento esercizio 4. Risolviamo (5) applicando (1):

$\begin{aligned} \lim_{n \to +\infty} \left( 1-\dfrac{1}{n^2}\right) ^n & = \lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{-n^2}\right) ^{-n^2}\right) ^{-\frac{1}{n}}=e^0=1 \end{aligned}$

Si conclude che:

$\boxcolorato{analisi}{\lim_{n \to +\infty} \left( 1-\dfrac{1}{n^2}\right) ^n=1 .}$

Esercizio 5. $(\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar)$ Calcolare il seguente limite, applicando (1):

(6) $\begin{equation*} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n}\right) ^{n^2}. \end{equation*}$

Svolgimento esercizio 5. Risolviamo (6) applicando (1):

$\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n}\right) ^{n^2} & = \lim_{n \to +\infty} \left( \underbrace{\left( 1+\dfrac{1}{n}\right) ^{n}}_{\to_{e}}\right) ^{\underbrace{n}_{\to{+\infty}}}=+\infty. \end{aligned}$

Si conclude che:

$\boxcolorato{analisi}{\lim_{n \to +\infty} \left( 1+\dfrac{1}{n}\right) ^{n^2}= +\infty. }$

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