Nepero numero 1

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Sia

    \[b_n:\mathbb{N}\setminus \{0\}\rightarrow \mathbb{R}\]

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:

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

 

Fonte: (clicca qui)