\begin{exercice*}[\ExerciceRefMethode{methode3-limA1}]
\label{exmethode3-limA1}
Déterminer les limites suivantes :
\begin{enumerate}
  \item $\displaystyle \lim_{n\rightarrow +\infty} n^2-2n$
  \item $\displaystyle \lim_{n\rightarrow +\infty} -3n^3+5n^2+6n-1$
  \item $\displaystyle \lim_{n\rightarrow +\infty} n^3-3n^4+2n^2-5n+2$
  \item $\displaystyle \lim_{n\rightarrow +\infty} (n^2-7n+2)(n^3-8n+1)$
  \item $\displaystyle \lim_{n\rightarrow +\infty} \dfrac{6n^2+3n+5}{-2n^2+5n-1}$
  \item $\displaystyle \lim_{n\rightarrow +\infty} \dfrac{9n^7-5n^4+n}{n^2+1}$
  \item $\displaystyle \lim_{n\rightarrow +\infty} \dfrac{6n^2+4n-2}{8n^3+7n^2-4n+7}$
\end{enumerate}

\begin{corrige}
\begin{enumerate}
\item
  $\displaystyle \lim_{n\rightarrow +\infty} n^2-2n=\lim_{n\rightarrow
    +\infty} n^2\left(1-\dfrac{2}{n}\right)=+\infty$

  \item $\displaystyle \lim_{n\rightarrow +\infty} -3n^3+5n^2+6n-1=-\infty$

\item
  $\displaystyle \lim_{n\rightarrow +\infty}
  n^3-3n^4+2n^2-5n+2=-\infty$

\item
  $\displaystyle \lim_{n\rightarrow +\infty}
  (n^2-7n+2)(n^3-8n+1)=+\infty$

\item
  $\displaystyle \lim_{n\rightarrow +\infty}
  \dfrac{6n^2+3n+5}{-2n^2+5n-1}= \lim_{n\rightarrow +\infty}
  \dfrac{n^2\left(6+\dfrac{3n}{n^2}+\dfrac{5}{n^2}\right)}{n^2\left(-2+\dfrac{5n}{n^2}-\dfrac{1}{n^2}\right)}
  = \lim_{n\rightarrow +\infty}
  \dfrac{6+\dfrac{3}{n}+\dfrac{5}{n^2}}{-2+\dfrac{5}{n}-\dfrac{1}{n^2}}$
  donc

  $\displaystyle \lim_{n\rightarrow +\infty}
  \dfrac{6n^2+3n+5}{-2n^2+5n-1}=-3.$

\item
  $\displaystyle \lim_{n\rightarrow +\infty}
  \dfrac{9n^7-5n^4+n}{n^2+1} = +\infty$

\item
  $\displaystyle \lim_{n\rightarrow +\infty}
  \dfrac{6n^2+4n-2}{8n^3+7n^2-4n+7}=0$
\end{enumerate}
  \end{corrige}
  \end{exercice*}

