87 lines
2.8 KiB
Markdown
87 lines
2.8 KiB
Markdown
#exercice
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title::"TD1 d'analyse"
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----
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sujet : [[L2_maths_analyse_TD1-annotate]]
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# Exercice 1
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Pour chacune des suites dont le terme général est donné ci-dessous, déterminer son éventuelle limite.
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Lorsque cette limite vaut $0^{+}$ ou $+\infty$, donner une équivalent simple.
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---
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$\disp u_{n} = \frac{n+\sqrt{n}}{n^{\frac{3}{2}}-n}$
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$$\begin{align*}
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u_{n} &= \frac{n + \sqrt{n}}{\sqrt{n}^{3}-n}\\
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&= \frac{n^{2} + n \sqrt{n}}{\sqrt{n} - 1}\\
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&= \frac{n^{2}\sqrt{n} + n^{2}}{1 - \frac{1}{\sqrt{n}}}\\
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\end{align*}$$
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$\lim\limits_{n \to \infty} \left( \frac{1-1}{\sqrt{n}} \right) = 1$
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$\lim\limits_{n \to \infty}\left( n^{2} \sqrt{n} + n^{2} \right) = +\infty$
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Donc, $\lim\limits_{n \to +\infty} u_{n} = +\infty$
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---
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$\disp u_{n} = \frac{2^{n}}{3^{n}+5^{n}}$
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$$\begin{align*}
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u_{n} &= \frac{2^{n}}{3^{n}+5^{n}}\\
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&= \frac{2^{n}}{5^{n}\left( 1+ \left( \frac{3}{5} \right)^{n} \right)}\\
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&= \left( \frac{2}{5} \right)^{n} \times \frac{1}{1+ \left(\frac{3}{5}\right)^{n}}
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\end{align*}$$
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$\lim\limits_{n \to +\infty} u_{n} = 0 \times \frac{1}{1+0}$ car $\frac{2}{5} < 1$ et $\frac{3}{5} < 1$
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Equivalent simple :
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puisque $\lim\limits_{n \to +\infty} \left( \dfrac{1}{1+\left( \frac{3}{5} \right)^{n}} \right) = 1$ et que $\lim\limits_{n \to +\infty} \left( \frac{2}{5} \right)^{n}$, alors on a :
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$u_{n} \sim_{+\infty} \left( \frac{2}{5} \right)^{n}$
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---
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$\disp u_{n} = \ln \left( 1+ \frac{1}{n} \right)$
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$\lim\limits_{n \to \infty} u_{n} = \ln(1+0) = 0$
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---
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$u_{n} = \sin \left( \frac{1}{n} \right)$
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$$\begin{align*}
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u_{n} &= \sin \left( \frac{1}{n} \right)\\
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&= X + \frac{X^{2}}{2} + o(X^{2}) \text{ où } X=\frac{1}{n}\\
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&= \frac{1}{n} + \frac{\frac{1}{n^{2}}}{2} + o\left(\frac{1}{n^{2}}\right)\\
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&= \frac{1}{n} + \frac{1}{2n^{2}} + o\left(\frac{1}{n^{2}}\right)\\
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\end{align*}$$
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Or, $\lim\limits_{n \to \infty} \left(\frac{1}{n}\right) = \lim\limits_{n \to \infty} \left(\frac{1}{2n^{2}}\right) = 0$
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Donc :
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$\lim\limits_{n \to \infty}u_{n} = 0$
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---
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$\disp u_{n} = e^{\frac{2n}{n+\sqrt{3n}}}$
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On étudie d'abord :
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$s_{n} = \frac{2n}{n+\sqrt{3n}}$
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$$\begin{align*}
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s_{n}&= \frac{2n}{n\left(1+ \frac{\sqrt{3n}}{\sqrt{n}}\right)}\\
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&= \frac{2}{1+ \frac{\sqrt{3}}{\sqrt{n}}}
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\end{align*}$$
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Donc :
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$\lim\limits_{n \to \infty} s_{n} = 2$
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et :
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$\lim\limits_{n \to \infty}u_{n} = \lim\limits_{n \to }e^{v_{n}} = e^{2}$
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---
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$$\begin{align*}
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u_{n} &= \tan \left(\frac{\pi}{2} \cos\left(\frac{1}{n}\right)\right)\\
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&= \tan\left(\frac{\pi}{2} \times 1^{-}\right)\\
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&= \tan \left( \frac{\pi}{2}^{-} \right)\\
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&= +\infty
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\end{align*}$$
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**Equivalent simple :**
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$$\begin{align*}
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u_{n} &= \tan \left( \frac{\pi}{2} \cos \left( \frac{1}{n} \right) \right)\\
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&= \tan \left( \frac{\pi}{2} \left( 1 - \frac{1}{2n^{2}} + o\left(\frac{1}{n^{2}}\right) \right) \right)\\
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&= \tan \left( \frac{\pi}{2} - \frac{\pi}{2n^{2}} + o \left( \frac{1}{n^{2}} \right) \right)\\
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&=
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\end{align*}$$
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