Barre utilisateur

[ID: 738] [Date de publication: 18 janvier 2021 13:47] [Catégorie(s): Applications au calcul de limite ] [ Nombre commentaires: 1] [nombre d'éditeurs: 1 ] [Editeur(s): Emmanuel Vieillard-Baron ] [nombre d'auteurs: 3 ] [Auteur(s): Emmanuel Vieillard-Baron Alain Soyeur François Capaces ]

Solution(s)

Solution(s)

Exercice 55
Par emmanuel le 18 janvier 2021 13:48

1. \[\begin{aligned} \dfrac{\ln\left(2x^2-1\right)}{\tan \left(x-1\right)}&\xlongequal{X=x-1}&\dfrac{\ln\left(1+4X+2X^2\right)}{\tan X}\\ &=& \dfrac{4X+\underset{X \rightarrow 0}{o}\left(X\right)}{X+\underset{X \rightarrow 0}{o}\left(X\right)}\\ &=& \dfrac{4+\underset{X \rightarrow 0}{o}\left(1\right)}{1+\underset{X \rightarrow 0}{o}\left(1\right)}\\ &\xrightarrow[X\rightarrow 0]{}&\boxed{4} \end{aligned}\]

2. \[\begin{aligned} x-x^2\ln\left(1+\dfrac{1}{x}\right)&\xlongequal{X={\scriptstyle 1\over\scriptstyle x}}&\dfrac{1}{X}-\dfrac{1}{X^2}\ln\left(1+X\right)\\ &=& \dfrac{1}{X}-\dfrac{1}{X^2}\left(X-{\scriptstyle X^2\over\scriptstyle 2}+\underset{X \rightarrow 0}{o}\left(X^2\right)\right)\\ &=&\dfrac{1}{2}+\underset{X \rightarrow 0}{o}\left(X\right)\\ &\xrightarrow[X\rightarrow 0]{}& \boxed{\dfrac{1}{2}} \end{aligned}\]

3. \[\begin{aligned} \dfrac{x-\operatorname{arctan} x}{\sin^3 x}&\underset{x\rightarrow 0}{\sim}&\dfrac{x-\operatorname{arctan} x}{x^3}\\ &=& \dfrac{{\scriptstyle 1\over\scriptstyle 3}x^3+\underset{x \rightarrow 0}{o}\left(x^3\right)}{x^3}\\ &=&{\scriptstyle 1\over\scriptstyle 3}+\underset{x \rightarrow 0}{o}\left(1\right)\\ &\xrightarrow[x\rightarrow 0]{}&\boxed{{\scriptstyle 1\over\scriptstyle 3}} \end{aligned}\]

4. \[\begin{aligned} \dfrac{1}{x^3}-\dfrac{1}{\sin^3 x}&=&\dfrac{\sin^3 x - x^3}{x^3\sin^3 x}\\ &\underset{x\rightarrow 0}{\sim}&\dfrac{\sin^3 x - x^3}{x^6}\\ &=&\dfrac{-{\scriptstyle x^5\over\scriptstyle 2}+\underset{x \rightarrow 0}{o}\left(x^6\right) }{x^6}\\ &=&-{\scriptstyle 1\over\scriptstyle 2x}+\underset{x \rightarrow 0}{o}\left(1\right) \end{aligned}\] et \(\boxed{\displaystyle{\lim_{x \rightarrow 0^+}\dfrac{1}{x^3}-\dfrac{1}{\sin^3 x}}=-\infty}\) tandis que \(\boxed{\displaystyle{\lim_{x \rightarrow 0^-}\dfrac{1}{x^3}-\dfrac{1}{\sin^3 x}}=+\infty}\)

5. \[\begin{aligned} \dfrac{1}{x}-\dfrac{1}{\ln(1+x)}&=&\dfrac{\ln\left(1+x\right)-x}{x\ln(1+x)}\\ &\underset{x\rightarrow 0}{\sim}&\dfrac{\ln\left(1+x\right)-x}{x^2}\\ &=&\dfrac{-{\scriptstyle x^2\over\scriptstyle 2}+\underset{x \rightarrow 0}{o}\left(x^2\right)}{x^2}\\ &=& -{\scriptstyle 1\over\scriptstyle 2}+\underset{x \rightarrow 0}{o}\left(1\right)\\ &\xrightarrow[x\rightarrow 0]{}-{\scriptstyle 1\over\scriptstyle 2} \end{aligned}\]

6. \[\begin{aligned} \left(\dfrac{1^x+2^x+\dots+n^x}{n}\right)^{{\scriptstyle 1\over\scriptstyle x}}&=&\left(\dfrac{ \left(1+x\ln 1 \right)+ \left(1+ x \ln 2\right)+\dots+ \left(1+x\ln n\right)+\underset{x \rightarrow 0}{o}\left(x\right) }{ n}\right)^{{\scriptstyle 1\over\scriptstyle x}}\\ &=& \left(\dfrac{ n+\left(\ln 1 + \dots+ \ln n\right)x+\underset{x \rightarrow 0}{o}\left(x\right) }{ n}\right)^{{\scriptstyle 1\over\scriptstyle x}}\\ &=& \left(1+\ln\left(n!^{{\scriptstyle 1\over\scriptstyle n}}\right)x+\underset{x \rightarrow 0}{o}\left(x\right)\right)^{{\scriptstyle 1\over\scriptstyle x}}\\ &=&e^{\dfrac{\ln \left(1+\ln\left(n!^{{\scriptstyle 1\over\scriptstyle n}}\right)x+\underset{x \rightarrow 0}{o}\left(x\right)\right) }{x}}\\ &=& e^{ \dfrac{\ln\left(n!^{{\scriptstyle 1\over\scriptstyle n}}\right)x+\underset{x \rightarrow 0}{o}\left(x\right)} {x} }\\ &=&e^{\left( \ln\left(n!^{{\scriptstyle 1\over\scriptstyle n}}\right)+\underset{x \rightarrow 0}{o}\left(1\right)\right)}\\ &\xrightarrow[x\rightarrow 0]{}&e^{ \ln\left(n!^{{\scriptstyle 1\over\scriptstyle n}}\right)}=\boxed{n!^{{\scriptstyle 1\over\scriptstyle n}}} \end{aligned}\]