Réponse :
Bonsoir,
Explications étape par étape :
[tex]\left\{\begin{array}{ccc}u_0&=&-\dfrac{91}{10} \\\\u_{n+1}&=&\dfrac{10u_n+81}{-u_n-8} \\\end{array}\right.\\\\\\v_n=\dfrac{1}{u_n+9} \\\\v_{n+1}=\dfrac{1}{u_{n+1}+9} \\=\dfrac{1}{\dfrac{10u_n+81}{-u_n-8} +9} \\\\\\=-\dfrac{u_n+8}{u_n+9} \\\\\\=-\dfrac{u_n+8+1-1}{u_n+9} \\\\=-1+\dfrac{1}{u_+9} \\\\\\\boxed{v_{n+1}=v_n-1}\\\\v_0=-10\\\\\\v_n=-10-n\\\\u_n=\dfrac{1}{-10-n} -9=-\dfrac{9n+91}{n+10} \\\\[/tex]
[tex]\displaystyle \lim_{n \to \infty} u_n =\lim_{n \to \infty} -\dfrac{9n+91}{n+10} =-9[/tex]
