DS2 2021 2022 Correction Analyse2-DS PREING1 S2
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Correction du DS2 d’Analyse 2 de 2021/2022 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 :
1.
$\sqrt{n^{4} + 1} - n^{2} = n^{2}\left( \sqrt{1 + \frac{1}{n^{4}}} - 1 \right)$
$\text{Or, avec}\ X \rightarrow 0,\ \left( \sqrt{1 + X} - 1 \right)\ \sim\frac{X}{2},\ \text{donc}\ \left( \sqrt{1 + \frac{1}{n^{4}}} - 1 \right)\ \sim\frac{1}{2n^{4}}$
$\text{Ainsi}\ \left( \sqrt{n^{4} + 1} - n^{2} \right)\ \sim\frac{1}{2n^{2}}$
$\text{Aussi}\text{, avec}\ X \rightarrow 0,\ (e^{X} - 1)\ \sim\ X,\ \text{donc}\ \left( \exp\left( \frac{( - 1)^{n}}{n} \right) - 1 \right)\ \sim\frac{( - 1)^{n}}{n}$
Par quotient,
$u_{n}\ \sim\frac{( - 1)^{n}}{2n}$
2.
$\ln\left( 1 + \sin(x) \right)\ \sim_{x \rightarrow 0}\sin(x)\ \sim_{x \rightarrow 0}\ x$
$\sin\left( \pi x^{2} \right)\ \sim_{0}\ \pi x^{2}$
$f(x)\ \sim_{x \rightarrow 0}\frac{1}{\pi x}$
3.
$g(x) = \frac{\sin(x)\left( \cos(x) - 1 \right)}{\left( x^{4} + x^{5} \right)\left( \cos(\pi x) \right)}$
$\sin(x)\ \sim_{0}\ x$
$\cos(x) - 1\ \sim_{0}\ \left( - \frac{x^{2}}{2} \right)$
$\left( x^{4} + x^{5} \right)\ \sim_{0}\ x^{4}$
$\cos(\pi x)\ \sim_{0}\ 1$
$g(x)\ \sim_{0}\ x\left( - \frac{x^{2}}{2} \right)\left( \frac{1}{x^{4}} \right) = - \frac{1}{2x}$
$g(x)\ \sim_{x \rightarrow 0}\left( - \frac{1}{2x} \right)$
Exercice 2 :
1.a.
$\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{8} + o\left( x^{2} \right)$
$\frac{1}{1 + x + x^{2}} = 1 - \left( x + x^{2} \right) + \left( x + x^{2} \right)^{2} + o\left( x^{2} \right) = 1 - x + o\left( x^{2} \right)$
$\frac{\sqrt{1 + x}}{1 + x + x^{2}} = \left( 1 + \frac{x}{2} - \frac{x^{2}}{8} \right)(1 - x) + o\left( x^{2} \right) = 1 - x + \frac{x}{2} - \frac{x^{2}}{2} - \frac{x^{2}}{8} + o\left( x^{2} \right)$
$g_{1}(x) = 1 - \frac{x}{2} - \frac{5x^{2}}{8} + o\left( x^{2} \right)$
1.b.
$f(x) = - \frac{5x^{2}}{8} + o\left( x^{2} \right)$
$f(x)\ \sim_{x \rightarrow 0} - \frac{5x^{2}}{8}$
2.
$g_{2}(x) = e^{\frac{\ln(1 + x)}{x}}$
$\ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + o\left( x^{3} \right)$
$\frac{\ln(1 + x)}{x} = 1 - \frac{x}{2} + \frac{x^{2}}{3} + o\left( x^{2} \right)$
$\text{On pose}\ X = - \frac{x}{2} + \frac{x^{2}}{3} + o\left( x^{2} \right)$
$e^{\frac{\ln(1 + x)}{x}} = e^{1 + X} = e\left( e^{X} \right) = e\left( 1 + X + \frac{X^{2}}{2} + o\left( X^{2} \right) \right)$
$g_{2}(x) = e\left( 1 + \left( - \frac{x}{2} + \frac{x^{2}}{3} \right) + \frac{1}{2}\left( - \frac{x}{2} + \frac{x^{2}}{3} \right)^{2} \right) + o\left( x^{2} \right)$
$g_{2}(x) = e\left( 1 - \frac{x}{2} + \frac{x^{2}}{3} + \frac{x^{2}}{8} \right) + o\left( x^{2} \right)$
$g_{2}(x) = e - \frac{e}{2}x + \frac{11e}{24}x^{2} + o\left( x^{2} \right)$
3.
Faire un DL de $g_{3}(x)$ au voisinage de $\frac{\pi}{2}$ revient à faire un DL de $g_{3}\left( x + \frac{\pi}{2} \right)$ en 0.
$g_{3}\left( x + \frac{\pi}{2} \right) = e^{x + \frac{\pi}{2}}\sin\left( x + \frac{\pi}{2} \right) = e^{\frac{\pi}{2}}e^{x}\cos(x)$
$e^{x} = 1 + x + \frac{x^{2}}{2} + o\left( x^{2} \right)$
$\cos(x) = 1 - \frac{x^{2}}{2} + o\left( x^{2} \right)$
$g_{3}\left( x + \frac{\pi}{2} \right) = e^{\frac{\pi}{2}}\left( 1 + x + \frac{x^{2}}{2} \right)\left( 1 - \frac{x^{2}}{2} \right) + o\left( x^{2} \right)$
$g_{3}\left( x + \frac{\pi}{2} \right) = e^{\frac{\pi}{2}}\left( 1 - \frac{x^{2}}{2} + x + \frac{x^{2}}{2} \right) + o\left( x^{2} \right)$
$g_{3}\left( x + \frac{\pi}{2} \right) = e^{\frac{\pi}{2}} + e^{\frac{\pi}{2}}x + o\left( x^{2} \right)$
$g_{3}(x) = e^{\frac{\pi}{2}} + e^{\frac{\pi}{2}}\left( x - \frac{\pi}{2} \right) + o\left( \left( x - \frac{\pi}{2} \right)^{2} \right)$
Exercice 3 :
1.
$\cos\left( \frac{1}{n^{2}} \right) - 1\ \sim - \frac{1}{2n^{4}}$
$n^{2}\left( \cos\left( \frac{1}{n^{2}} \right) - 1 \right)\ \sim - \frac{1}{2n^{2}}$
$\lim_{n \rightarrow + \infty}{\ln\left( 1 + \frac{1}{n} \right)} = 0,\ \text{donc}\sin\left( \ln\left( 1 + \frac{1}{n} \right) \right)\ \sim\ln\left( 1 + \frac{1}{n} \right)\ \sim\frac{1}{n}$
$\frac{n^{2}\left( \cos\left( \frac{1}{n^{2}} \right) - 1 \right)}{\sin\left( \ln\left( 1 + \frac{1}{n} \right) \right)}\ \sim - \frac{1}{2n}$
$\lim_{n \rightarrow + \infty}\frac{n^{2}\left( \cos\left( \frac{1}{n^{2}} \right) - 1 \right)}{\sin\left( \ln\left( 1 + \frac{1}{n} \right) \right)} = 0$
2.
$\sin\left( \frac{\pi}{n} \right)\ \sim\frac{\pi}{n}$
$\ln\left( 1 + \frac{2}{n} \right)\ \sim\frac{2}{n}$
$\ln\left( 1 + \frac{3}{n} \right)\ \sim\frac{3}{n}$
$\frac{n\sin\left( \frac{\pi}{n} \right)\ln\left( 1 + \frac{2}{n} \right)}{\ln\left( 1 + \frac{3}{n} \right)}\ \sim\frac{2\pi}{3}$
$\lim_{n \rightarrow + \infty}\frac{n\sin\left( \frac{\pi}{n} \right)\ln\left( 1 + \frac{2}{n} \right)}{\ln\left( 1 + \frac{3}{n} \right)} = \frac{2\pi}{3}$
3.
$\sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + o\left( x^{5} \right)$
$\cos(x) = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} + o\left( x^{4} \right)$
$3\sin(x) - x\cos(x) - 2x = 3x - \frac{x^{3}}{2} + \frac{x^{5}}{40} - x + \frac{x^{3}}{2} - \frac{x^{5}}{24} - 2x + o\left( x^{5} \right)$
$3\sin(x) - x\cos(x) - 2x = x^{5}\left( \frac{1}{40} - \frac{1}{24} \right) = - \frac{x^{5}}{60} + o\left( x^{5} \right) = x^{5}\left( - \frac{1}{60} + o(1) \right)$
$\pi x^{5} + x^{6} = x^{5}(\pi + x)$
$\frac{3\sin(x) - x\cos(x) - 2x}{\pi x^{5} + x^{6}} = \frac{- \frac{1}{60} + o(1)}{\pi + x}$
$\lim_{x \rightarrow 0}\frac{3\sin(x) - x\cos(x) - 2x}{\pi x^{5} + x^{6}} = - \frac{1}{60\pi}$
4.
$\ln(1 + 2x) = 2x - \frac{(2x)^{2}}{2} + o\left( x^{2} \right) = 2x - 2x^{2} + o\left( x^{2} \right)$
$e^{2x} = 1 + 2x + \frac{(2x)^{2}}{2} = 1 + 2x + 2x^{2} + o\left( x^{2} \right)$
$1 - \cos(x) = \frac{x^{2}}{2} + o\left( x^{2} \right)$
$1 + \ln(1 + 2x) - e^{2x} = - 4x^{2} + o\left( x^{2} \right)$
$\frac{1 + \ln(1 + 2x) - e^{2x}}{1 - \cos(x)} = \frac{- 4x^{2} + o\left( x^{2} \right)}{\frac{x^{2}}{2} + o\left( x^{2} \right)} = \frac{- 4 + o(1)}{\frac{1}{2} + o(1)}$
$\lim_{x \rightarrow 0}\frac{1 + \ln(1 + 2x) - e^{2x}}{1 - \cos(x)} = - 8$
Exercice 4 :
1.
$f(x) = \sqrt{\frac{\left( \frac{1}{v} \right)^{3}}{\frac{1}{v} + 1}} = \sqrt{\frac{1}{v^{2} + v^{3}}}$
$f(x) = \sqrt{\frac{1}{v^{2}(1 + v)}}$
$f(x) = \frac{1}{v}\sqrt{\frac{1}{1 + v}}$
$f(x) = \frac{1}{v}\sqrt{1 - v + v^{2} + o_{v \rightarrow 0}\left( v^{2} \right)}$
Soit $u = - v + v^{2} + o_{v \rightarrow 0}\left( v^{2} \right)$
$\sqrt{1 + u} = 1 + \frac{u}{2} - \frac{u^{2}}{8} + o_{u \rightarrow 0}\left( u^{2} \right)$
$\sqrt{1 - v + v^{2} + o_{v \rightarrow 0}\left( v^{2} \right)} = 1 + \frac{1}{2}\left( - v + v^{2} \right) - \frac{1}{8}\left( - v + v^{2} \right)^{2} + o_{v \rightarrow 0}\left( v^{2} \right)$
$\sqrt{1 - v + v^{2} + o_{v \rightarrow 0}\left( v^{2} \right)} = 1 - \frac{v}{2} + \frac{3}{8}v^{2} + o_{v \rightarrow 0}\left( v^{2} \right)$
$f(x) = \frac{1}{v} - \frac{1}{2} + \frac{3}{8}v + o_{v \rightarrow 0}(v)$
$f(x) = x - \frac{1}{2} + \frac{3}{8x} + o_{x \rightarrow + \infty}\left( \frac{1}{x} \right)$
2.
Une asymptote oblique de $C_{f}$ en $+ \infty$ est de forme :
$y = x - \frac{1}{2}$
3.
$f(x) - \left( x - \frac{1}{2} \right) = \frac{3}{8x} + o\left( \frac{1}{x} \right)\ \sim\frac{3}{8x} > 0$
Donc $C_{f}$ est au-dessus de l’asymptote oblique.