Nepero numero 2

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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*}

    \[\ast \quad \ast \quad \ast\]

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

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

 

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

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

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1-\dfrac{1}{n}\right) ^{n^2}=0.}\]

 


Esercizio 2. (\bigstar\bigstar\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):

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

Si conclude che:

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

 

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

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

 

 

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

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

Si conclude che:

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

 

 

 

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^4}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^{n^4} & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{n^2}\right) ^{n^2}\right) ^{n^2}=e^{+\infty}=+\infty. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^{n^4}=+\infty .}\]

 

 

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^3}\right) ^{n^2}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( 1-\dfrac{1}{n^3}\right) ^{n^2} & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{-n^3}\right) ^{-n^3}\right) ^{\dfrac{n^2}{-n^3}}=e^{0}=1. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1-\dfrac{1}{n^3}\right) ^{n^2} =1 .}\]

 

Fonte: (clicca qui)