Nepero numero 3

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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^2}\right) ^{2n^2+5}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^{2n^2+5} & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{n^2}\right) ^{n^2}\right) ^{\dfrac{2n^2\left( 1+\frac{5}{2n^2}\right) }{n^2}}=e^{2}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^2}\right) ^{2n^2+5}=e^2.}\]

 


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

(3)   \begin{equation*} \lim_{n \to +\infty} \left( 1+\dfrac{7}{n}\right)^{\frac{3n^2+5}{4n+1}}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{7}{n}\right) ^{\dfrac{3n^2+5}{4n+1}} & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{\frac{n}{7}}\right) ^{\dfrac{n}{7}}\right) ^{\dfrac{7}{n}\cdot\dfrac{3n^2\left( 1+\frac{5}{3n^2}\right) }{4n\left( 1+\frac{1}{4n}\right) }}=e^{\frac{21}{4}}=\sqrt[4]{e^{21}}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1+\dfrac{7}{n}\right) ^{\frac{3n^2+5}{4n+1}} =\sqrt[4]{e^{21}}.}\]

 

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^2}\right) ^{\frac{n^4+n+1}{n^2+7}}. \end{equation*}

 

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( 1-\dfrac{1}{n^2}\right) ^{\dfrac{n^4+n+1}{n^2+7}} & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{-n^2 }\right) ^{-n^2} \right)^{\dfrac{n^4\left( 1+\frac{1}{n^3}+\frac{1}{n^4}\right) }{-n^4\left( 1+\frac{7}{n^2}\right) }}=e^{-1}=\dfrac{1}{e}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \left( 1-\dfrac{1}{n^2}\right) ^{\frac{n^4+n+1}{n^2+7}}=\dfrac{1}{e}.}\]

 

 

 

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

(5)   \begin{equation*} \lim_{n \to +\infty} \left( \dfrac{n+9}{n+5}\right) ^{3n}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( \dfrac{n+9}{n+5}\right) ^{3n} & = \lim_{n \to +\infty} \left( \dfrac{n+5+4}{n+5}\right) ^{3n} =\lim_{n \to +\infty}\left( 1+\dfrac{4}{n+5}\right) ^{3n}=\\\\ & =\lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{\dfrac{n+5}{4}}\right) ^{\dfrac{n+5}{4}}\right)^{\dfrac{12n}{n\left( 1+\frac{5}{n}\right) }}=e^{12}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \left( \dfrac{n+9}{n+5}\right) ^{3n}=e^{12} .}\]

 

 

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

(6)   \begin{equation*} \lim_{n \to +\infty} \left( \dfrac{n+5}{n+8}\right) ^{\frac{n}{4}} \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( \dfrac{n+5}{n+8}\right)^{\frac{n}{4}} & = \lim_{n \to +\infty} \left( \dfrac{n+8-3}{n+8}\right) ^{\frac{n}{4}}= \lim_{n \to +\infty} \left( 1-\dfrac{3}{n+8}\right)^{\frac{n}{4}}=\\\\ & =\lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{-\dfrac{n+8}{3}}\right) ^{-\dfrac{n+8}{3}}\right)^{-\dfrac{3n}{4n\left( 1+\frac{8}{n}\right) }}=e^{-\frac{3}{4}}=\dfrac{1}{\sqrt[4]{e^3}}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( \dfrac{n+5}{n+8}\right) ^{\frac{n}{4}}=\dfrac{1}{\sqrt[4]{e^3}} .}\]

 

Fonte: (clicca qui)