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\title{Chapitre 5 : Les Limites }
\author{D. Zancanaro \and C. Aup\'erin}
\date{2009-2010}

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\begin{document}

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{\Large Lyc\'ee Jean Durand, Castelnaudary\hfill
Classe de $1^{\text{\`e}re}$ S}\par

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\vfill\vfill
\null\hfill ``T\'el\'echarger c'est tuer l'industrie, tuons les tous'' \textsc{Thurston Moore}\\
\null\hfill Derni\`ere modification : \today
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\lhead{Chapitre 5}
\rhead{Les Limites}

\tableofcontents 
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\Cv{Cours : Les Limites des fonctions de r\'ef\'erences}

\section{Fonction carr\'ee}

\begin{multicols}{2}
On consid\`ere la fonction d\'efinie sur $\R$ \`a valeur dans $\R^+$ par $ f(x)=x^2 $. On a alors
$$ \lim\limits_{x \to +\infty} f(x)=+\infty$$
$$\lim\limits_{x \to -\infty} f(x)=+\infty$$
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\end{multicols}

\section{Fonction cube}
\begin{multicols}{2}
On consid\`ere la fonction d\'efinie sur $\R$ \`a valeur dans $\R$ par $ f(x)=x^3 $. On a alors
$$ \lim\limits_{x \to +\infty} f(x)=+\infty$$
$$\lim\limits_{x \to -\infty} f(x)=-\infty$$
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\section{Fonction racine carr\'ee}
\begin{multicols}{2}
On consid\`ere la fonction d\'efinie sur $\R^+$ \`a valeur dans $\R^+$ par $ f(x)=\sqrt{x} $. On a alors
$$ \lim\limits_{x \to 0^+} f(x)=0$$
$$\lim\limits_{x \to +\infty} f(x)=+\infty$$
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%\newpage

\begin{multicols}{2}
\section{Fonction inverse}
On consid\`ere la fonction d\'efinie sur $\R-\{0\}$ \`a valeur dans $\R-\{0\}$ par $ f(x)=\dfrac{1}{x} $. On a alors
$$ \lim\limits_{x \to -\infty} f(x)=0 \qquad \qquad \lim\limits_{x \to +\infty} f(x)=0$$
$$\lim\limits_{x \to 0^{-}} f(x)=-\infty \qquad \qquad \lim\limits_{x \to 0^{+}} f(x)=+\infty$$
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\section{Fonctions Puissances}
On consid\`ere la fonction d\'efinie sur $\R$ par $f(x)=x^n$ o\`u 
$n\in\Z$
\begin{itemize}
\item Si $n>0$ : $\lim\limits_{x \to +\infty} f(x)=+\infty \qquad \quad$ et 
$ \qquad\quad \lim\limits_{x \to -\infty} f(x)= 
\left\{   \begin{array}{ll}
           +\infty \qquad \text{ si n est pair} \\ -\infty\qquad \text{ si n est impair}
          \end{array}\right.$ \\

\item Si $n<0$ : $\lim\limits_{x \to +\infty} f(x)=0 \qquad \qquad \, $ et $\qquad \quad \lim\limits_{x \to -\infty} f(x)=0$ \\

\Expls{\begin{itemize}
\item $ \lim\limits_{x \to +\infty} x^4=+\infty \qquad $ et $ \qquad \lim\limits_{x \to -\infty} x^4=+\infty$ \\

\item $\lim\limits_{x \to +\infty} x^5=+\infty \qquad$ et $\qquad \lim\limits_{x \to -\infty} x^5=-\infty$\\

\item $\lim\limits_{x \to +\infty} x^{-3}=\lim\limits_{x \to +\infty} \dfrac{1}{x^{3}} =0  \qquad \quad $ 
et $\qquad \lim\limits_{x \to -\infty} x^{-3}=\lim\limits_{x \to -\infty} \dfrac{1}{x^{3}} =0  $\\
 
\item $\lim\limits_{x \to -\infty} x^{-10}=\lim\limits_{x \to -\infty} \dfrac{1}{x^{10}} =0 \qquad \, $
et $\qquad \lim\limits_{x \to -\infty} x^{-10}=\lim\limits_{x \to -\infty} \dfrac{1}{x^{10}} =0  $
\end{itemize}
}
\end{itemize}

\section{Fonctions polyn\^omes}
On consid\`ere la fonction $f$ d\'efinie sur $\R$ par $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0$ o\`u $a_n\neq 0$
\begin{eqnarray*}
&& \lim\limits_{x \to +\infty} f(x)= \lim\limits_{x \to -\infty} a_nx^n=\left\lbrace
          \begin{array}{ll}
            +\infty & \qquad \mathrm{si}\quad a_n> 0 \\
            -\infty & \qquad \mathrm{si} \quad a_n< 0\\
          \end{array}
        \right. \\\\
&& \lim\limits_{x \to -\infty} f(x)= \lim\limits_{x \to -\infty} a_nx^n=\pm \infty
\end{eqnarray*}

\Expls{\begin{itemize}
\item $\lim\limits_{x \to -\infty} (-7x^3+2x^2-4x+2)=\lim\limits_{x \to -\infty} -7x^3 =+\infty$\\

\item $\lim\limits_{x \to -\infty} (2x^3+2x^2-4x+2)=\lim\limits_{x \to -\infty} 2x^3 =-\infty$
       \end{itemize}
}


\section{Fonctions rationnelles}
Soit la fonction $f$ d\'efinie par $f(x)=\dfrac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_rx^r+b_{r-1}x^{r-1}+\dots+b_2x^2+b_1x+b_0}$ o\`u $a_n\neq 0$ et $b_r\neq 0$\\

$$\lim\limits_{x \to +\infty} f(x)= \lim\limits_{x \to +\infty} \dfrac{a_nx^n}{b_rx^r} \qquad 
\textrm{ et } \qquad \lim\limits_{x \to -\infty} f(x)= \lim\limits_{x \to -\infty} \dfrac{a_nx^n}{b_rx^r}$$\\
\Expls{\begin{itemize}
\item $\lim\limits_{x \to +\infty} \dfrac{2x^2-4x-6}{x-2} = \lim\limits_{x \to +\infty} \dfrac{2x^2}{x}=\lim\limits_{x \to +\infty} 2x=+\infty$\\

\item $\lim\limits_{x \to -\infty} \dfrac{-x^3-4x-6}{2x^3+x-2} = \lim\limits_{x \to -\infty} \dfrac{-x^3}{2x^3}=\lim\limits_{x \to -\infty} -\dfrac{1}{2} =-\dfrac{1}{2}$ \\

\item $\lim\limits_{x \to -\infty} \dfrac{3x-1}{2x^3+x-2} = \lim\limits_{x \to -\infty} \dfrac{3x}{2x^3}=\lim\limits_{x \to -\infty} \dfrac{3}{2x^2} =0$
       \end{itemize}
}

\end{document}