Bonjour Macklewis
[tex]1)\ u_n=2n^3-n^2+1\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}2n^3(1-\dfrac{n^2}{2n^3}+\dfrac{1}{2n^3})\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}2n^3(1-\dfrac{1}{2n}+\dfrac{1}{2n^3})\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}2n^3\times\lim\limits_{n\to+\infty}(1-\dfrac{1}{2n}+\dfrac{1}{2n^3})[/tex]
Or
[tex]\lim\limits_{n\to+\infty}2n^3=+\infty\\\lim\limits_{n\to+\infty}(1-\dfrac{1}{2n}+\dfrac{1}{2n^3})=\lim\limits_{n\to+\infty}1-\lim\limits_{n\to+\infty}\dfrac{1}{2n}+\lim\limits_{n\to+\infty}\dfrac{1}{2n^3}\\\\=1-0+1=1[/tex]
D'où
[tex]\lim\limits_{n\to+\infty}u_n=+\infty\times1\\\\\boxed{\lim\limits_{n\to+\infty}u_n=+\infty}[/tex]
[tex]3)\ u_n=\dfrac{-n+1}{2n^2-4}\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}\dfrac{-n(1+\dfrac{1}{-n})}{2n^2(1-\dfrac{4}{2n^2})}\\\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}\dfrac{-n(1-\dfrac{1}{n})}{2n^2(1-\dfrac{2}{n^2})}\\\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}\dfrac{-(1-\dfrac{1}{n})}{2n(1-\dfrac{2}{n^2})}\\\\\\\lim\limits_{n\to+\infty}u_n=\lim\limits_{n\to+\infty}\dfrac{-1}{2n}\times\dfrac{1-\dfrac{1}{n}}{1-\dfrac{2}{n^2}}[/tex]
Or
[tex]\lim\limits_{n\to+\infty}\dfrac{-1}{2n}=0\\\\\lim\limits_{n\to+\infty}\dfrac{1-\dfrac{1}{n}}{1-\dfrac{2}{n^2}}=\dfrac{1-0}{1-0}\Longrightarrow\lim\limits_{n\to+\infty}\dfrac{1-\dfrac{1}{n}}{1-\dfrac{2}{n^2}}=1[/tex]
Par conséquent,
[tex]\lim\limits_{n\to+\infty}u_n=0\times1\\\\\boxed{\lim\limits_{n\to+\infty}u_n=0}[/tex]
