DS2 2023 2024 V1 Correction Analyse2-DS PREING1 S2
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Correction du DS2 d’Analyse 2 de 2023/2024 par Mathis S.
La plupart des résultats ont été vérifiés par l’outil informatique, mais si vous constatez une erreur, merci de contacter Mathis S.
Exercice 1 :
1a.
$g(x) = \frac{\left( 1 - \cos(x) \right)\ln\left( 1 + ex^{3} \right)}{x^{3}\tan(x)}$
$1 - \cos(x)\ \sim\frac{x^{2}}{2}$
$\ln\left( 1 + ex^{3} \right)\ \sim\ ex^{3}$
$\tan(x)\ \sim\ x$
$g(x)\ \sim\frac{ex}{2}$
1b.
$h(x) = \frac{\cos x - 1}{e^{x} - 1}$
$\cos(x) - 1\ \sim - \frac{x^{2}}{2}$
$e^{x} - 1\ \sim\ x$
$h(x)\ \sim - \frac{x}{2}$
2
$\lim_{x \rightarrow 0}\frac{g(x)}{h(x)} = \lim_{x \rightarrow 0}{\left( \frac{ex}{2} \right)\left( - \frac{2}{x} \right)} = \lim_{x \rightarrow 0}{- e} = - e$
Exercice 2 :
1
$x\ln(1 - x) = x\left( ( - x) - \frac{( - x)^{2}}{2} + \frac{( - x)^{3}}{3} + o\left( x^{3} \right) \right)$
$x\ln(1 - x) = x\left( - x - \frac{x^{2}}{2} - \frac{x^{3}}{3} + o\left( x^{3} \right) \right)$
$x\ln(1 - x) = - x^{2} - \frac{x^{3}}{2} - \frac{x^{4}}{3} + o\left( x^{4} \right)$
$\ln\left( \cos(x) \right) = \ln\left( 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} + o\left( x^{4} \right) \right)$
$\text{Soit}\ u = - \frac{x^{2}}{2} + \frac{x^{4}}{4!} + o\left( x^{4} \right)$
$\ln\left( \cos(x) \right) = u - \frac{u^{2}}{2} + o\left( u^{2} \right)$
$\ln\left( \cos(x) \right) = - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - \frac{x^{4}}{8} + o\left( x^{4} \right)$
$\ln\left( \cos(x) \right) = - \frac{x^{2}}{2} - \frac{x^{4}}{12} + o\left( x^{4} \right)$
$\frac{\ln\left( \cos(x) \right)}{x\ln(1 - x)} = \frac{- \frac{x^{2}}{2} - \frac{x^{4}}{12} + o\left( x^{4} \right)}{- x^{2} - \frac{x^{3}}{2} - \frac{x^{4}}{3} + o\left( x^{4} \right)}$
$f(x) = \frac{\left( - \frac{1}{2} - \frac{x^{2}}{12} + o\left( x^{2} \right) \right)}{- 1 - \frac{x}{2} - \frac{x^{2}}{3} + o\left( x^{2} \right)} = \frac{\left( \frac{1}{2} + \frac{x^{2}}{12} + o\left( x^{2} \right) \right)}{1 + \frac{x}{2} + \frac{x^{2}}{3} + o\left( x^{2} \right)}$
$\text{Soit}\ v = \frac{x}{2} + \frac{x^{2}}{3} + o\left( x^{2} \right)$
$\frac{1}{1 + v} = 1 - v + v^{2} + o\left( v^{2} \right)$
$\frac{1}{1 + v} = 1 - \frac{x}{2} - \frac{x^{2}}{3} + \frac{x^{2}}{4} + o\left( x^{2} \right) = 1 - \frac{x}{2} - \frac{x^{2}}{12} + o\left( x^{2} \right)$
$f(x) = \left( \frac{1}{2} + \frac{x^{2}}{12} + o\left( x^{2} \right) \right)\left( 1 - \frac{x}{2} - \frac{x^{2}}{12} + o\left( x^{2} \right) \right)$
$f(x) = \frac{1}{2} - \frac{x}{4} - \frac{x^{2}}{24} + \frac{2x^{2}}{24} + o\left( x^{2} \right)$
$f(x) = \frac{1}{2} - \frac{x}{4} + \frac{x^{2}}{24} + o\left( x^{2} \right)$
2
$\sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + o\left( x^{5} \right)$
$\sin^{2}(x) = x^{2} - \frac{2x^{4}}{3!} + o\left( x^{5} \right)$
$e^{x^{2}} = 1 + x^{2} + \frac{x^{4}}{2} + o\left( x^{5} \right)$
$e^{x^{2}} - 1 = x^{2} + \frac{x^{4}}{2} + o\left( x^{5} \right)$
$g(x) = \left( x^{2} - \frac{2x^{4}}{3!} + o\left( x^{5} \right) \right)\left( x^{2} + \frac{x^{4}}{2} + o\left( x^{5} \right) \right)$
$g(x) = x^{4} + o\left( x^{5} \right)$
Exercice 3 :
1
$f\ \sim_{x_{0}}g\ \text{donc}\lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)} = 1$
$g\ \sim_{x_{0}}h\ \text{donc}\lim_{x \rightarrow x_{0}}\frac{g(x)}{h(x)} = 1$
$\lim_{x \rightarrow x_{0}}\frac{f(x)}{h(x)} = \lim_{x \rightarrow x_{0}}{\frac{f(x)}{g(x)} \times \frac{g(x)}{h(x)}} = 1$
Donc $f\ \sim_{x_{0}}h$
2
$f(x) = o\left( x^{n} \right)$
$\lim_{x \rightarrow 0}\frac{f(x)}{x^{n}} = 0$
$m \leq n \Rightarrow n = m + p\ \text{où}\ p \geq 0$
$\lim_{x \rightarrow 0}\frac{f(x)}{x^{m}} = \lim_{x \rightarrow 0}\frac{f(x)}{x^{n - p}} = \lim_{x \rightarrow 0}\frac{f(x)x^{p}}{x^{n}} = \lim_{x \rightarrow 0}\frac{f(x)}{x^{n}} \times \lim_{x \rightarrow 0}x^{p} = 0 \times \lim_{x \rightarrow 0}x^{p} = 0$
$\text{Donc}\ f(x) = o\left( x^{m} \right)$
3
Soit $u = \frac{1}{x}$
$\lim_{x \rightarrow + \infty}\left( \frac{\sqrt{1 + x}}{x} \right) = \lim_{x \rightarrow + \infty}\frac{\sqrt{x\left( 1 + \frac{1}{x} \right)}}{x} = \lim_{x \rightarrow + \infty}{\left( \frac{\sqrt{x}}{x} \right)\sqrt{1 + \frac{1}{x}}} = \lim_{x \rightarrow + \infty}{\left( \frac{1}{\sqrt{x}} \right)\sqrt{1 + \frac{1}{x}}} = 0$
$\text{Donc}\ \sqrt{1 + x} = o(x)$
$\ln\left( x + \sqrt{1 + x} \right) = \ln\left( x + o_{+ \infty}(x) \right) = \ln\left( x\left( 1 + o_{+ \infty}(1) \right) \right) = \ln(x) + \ln\left( 1 + o_{+ \infty}(1) \right)$
$\lim_{x \rightarrow + \infty}\frac{\ln\left( x + \sqrt{1 + x} \right)}{\ln(x)} = \lim_{x \rightarrow + \infty}\frac{\ln(x) + \ln\left( 1 + o_{+ \infty}(1) \right)}{\ln(x)} = \lim_{x \rightarrow + \infty}{1 + \frac{\ln\left( 1 + o_{+ \infty}(1) \right)}{\ln(x)}} = 1$
$\text{Donc }\ln\left( x + \sqrt{1 + x} \right)\sim\ln(x)$
Exercice 4 :
1
$\forall n \geq 2,\ \ \ln(n + 1)u_{n} - \ln(n)u_{n - 1} = n$
$u_{1} = 0$
$2 = \ln(3)u_{2} - \ln(2)u_{1}$
$3 = \ln(4)u_{3} - \ln(3)u_{2}$
$n - 1 = \ln(n)u_{n - 1} - \ln(n - 1)u_{n - 2}$
$n = \ln(n + 1)u_{n} - \ln(n)u_{n - 1}$
Par somme,
$\sum_{k = 2}^{n}k = \ln(n + 1)u_{n} - \ln(2)u_{1}$
$\sum_{k = 2}^{n}k = \ln(n + 1)u_{n}$
$u_{n} = \left( \frac{1}{\ln(n + 1)} \right)\sum_{k = 2}^{n}k$
$\sum_{k = 2}^{n}k = \frac{n(n + 1)}{2} - 1\ \sim_{+ \infty}\frac{n^{2}}{2}$
$\ln(n + 1)\ \sim_{+ \infty}\ln(n)$
$u_{n}\ \sim_{+ \infty}\frac{n^{2}}{2\ln(n)}$
2
$\left( \begin{array}{r} n
5 \end{array} \right) = \frac{n!}{5!(n - 5)!} = \frac{n(n - 1)(n - 2)(n - 3)(n - 4)}{5!}$
$n(n - 1)(n - 2)(n - 3)(n - 4) = n^{5} + o_{+ \infty}\left( n^{5} \right)$
$n(n - 1)(n - 2)(n - 3)(n - 4)\ \sim_{+ \infty}\ n^{5}$
$5! = 120$
$\left( \begin{array}{r} n
5 \end{array} \right)\sim_{+ \infty}\frac{n^{5}}{120}$
3
$v_{n} = 4n\left( 1 - \cos\left( \frac{1}{n^{3}} \right) \right)$
$1 - \cos(X)\ \sim_{0}\frac{X^{2}}{2}$
$1 - \cos\left( \frac{1}{n^{3}} \right)\ \sim_{+ \infty}\frac{1}{2n^{6}}$
$v_{n}\sim_{+ \infty}\frac{2}{n^{5}}$
4
$\lim_{n \rightarrow + \infty}{\left( \begin{array}{r} n
5 \end{array} \right)v_{n}}\ $
$\left( \begin{array}{r} n
5 \end{array} \right)v_{n}\ \sim\frac{1}{60}$
$\lim_{n \rightarrow + \infty}{\left( \begin{array}{r} n
5 \end{array} \right)v_{n}} = \frac{1}{60}$