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Nepero numero 4

Limiti notevoli

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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*}

    \[\ast \quad \ast \quad \ast\]

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

(2)   \begin{equation*} \lim_{n \to +\infty} \left( \dfrac{n^2+2n+8}{n^2-3n+7}\right) ^{n}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( \dfrac{n^2+2n+8}{n^2-3n+7}\right) ^{n} & = \lim_{n \to +\infty} \left( \dfrac{n^2-3n+7+3n-7+2n+8}{n^2-3n+7}\right) ^{n}=\\\\ &=\lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{\dfrac{n^2-3n+7}{5n+1}}\right) ^{\dfrac{n^2-3n+7}{5n+1}}\right) ^{\dfrac{5n^2\left( 1+\frac{1}{5n}\right) }{n^2\left( 1-\frac{3}{n}+\frac{7}{n^2}\right) }}=e^{5}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( \dfrac{n^2+2n+8}{n^2-3n+7}\right) ^{n}=e^5 .}\]

 


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

(3)   \begin{equation*} \lim_{n \to +\infty} \left( \dfrac{n^2+n-6}{n^2+n+7}\right) ^{n}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \left( \dfrac{n^2+n-6}{n^2+n+7}\right) ^{n} & = \lim_{n \to +\infty} \left( \dfrac{n^2+n+7-7-6}{n^2+n+7}\right) ^{n}=\\\\ & =\lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{-\dfrac{n^2+n+7}{13}}\right) ^{-\dfrac{n^2+n+7}{13}}\right) ^{-\dfrac{13n}{n^2\left(1+\frac{1}{n}+\frac{7}{n^2} \right) }}=\\\\ & = e^{0}=1. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( \dfrac{n^2+n-6}{n^2+n+7}\right) ^{n}= 1.}\]

 

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

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

 

 

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

    \[\begin{aligned} &\lim_{n \to +\infty} \left( \dfrac{n^2-2n+1}{n^2-4n+7}\right) ^{\dfrac{n^4-n+3}{2n^2+5}} = \lim_{n \to +\infty} \left( \dfrac{n^2-4n+7+4n-7-2n+1}{n^2-4n+7}\right) ^{\dfrac{n^4-n+3}{2n^2+5}}=\\\\ &=\lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{\dfrac{n^2-4n+7}{2n-6} }\right) ^{\dfrac{n^2-4n+7}{2n-6} } \right) ^{\dfrac{2n-6}{n^2-4n+7}\cdot\dfrac{n^4-n+3}{2n^2+5}}=e^{+\infty}=+\infty. \end{aligned}\]

Si conclude che:

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

 

 

 

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

(5)   \begin{equation*} \lim_{n \to +\infty} \left( \dfrac{n^2-n-6}{n^2-n+2}\right) ^{\frac{n^4+n+3}{n^2-1}}. \end{equation*}

 

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

    \[\begin{aligned} & \lim_{n \to +\infty} \left( \dfrac{n^2-n-6}{n^2-n+2}\right) ^{\dfrac{n^4+n+3}{n^2-1}} = \lim_{n \to +\infty} \left( \dfrac{n^2-n+2-2-6}{n^2-n+2}\right) ^{\dfrac{n^4+n+3}{n^2-1}}=\\\\ &=\lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{-\dfrac{n^2-n+2}{8} }\right) ^{-\dfrac{n^2-n+2}{8} } \right) ^{-\dfrac{8}{n^2-n+2}\cdot\dfrac{n^4+n+3}{n^2-1}}=e^{-8}=\dfrac{1}{e^8}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \left( \dfrac{n^2-n-6}{n^2-n+2}\right) ^{\frac{n^4+n+3}{n^2-1}}=\dfrac{1}{e^8}.}\]

 

 

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

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

 

Svolgimento esercizio 5.   Ricordiamo

(7)   \begin{equation*} \log_a b=\dfrac{\log_c b}{\log_c a} \qquad \text{con } \; a,c>0 \; \wedge \; a,c \neq 1 \; \wedge \; b>0 \end{equation*}

e risolviamo (6) applicando (7) e (1):

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

Si conclude che:

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

 

Fonte: (clicca qui)