coucou
1) Un = 7n² - 4(-1)ⁿ - (6/(4n+7))
-1 ≤ (-1)ⁿ ≤ 1
4 ≥ -4 * (-1)ⁿ ≥ -4
7n² + 4 -(6/(4n+7)) ≥ 7n² -4*(-1)ⁿ-(6/(4n+7)) ≥ 7n² -4 -(6/(4n+7))
lim n tend vers +infini 7n²-4-(6/(4n+7)) = +infini
donc d apres le theoreme de minoration, lim n tend vers +infini Un = +infini
2) Vn = (√n - 5n ) /( n² - 7√n)
Vn = n(√n/n -5)/n(n-7√n/n) = (√n/n -5)/(n-7√n/n)
lim n tend +infini √n/n -5 = -5
lim n tend +infini n-7√n/n = +infini
par quotient lim n tend +infini Vn = 0
3) Vn = (3n -sin n)/(5n² + 4)
-1 ≤ sin n ≤ 1
1 ≥ -sin n ≥ -1
3n +1 ≥ 3n -sin n ≥ 3n -1
(3n+1)/(5n²+4) ≥ (3n-sin n)/(5n²+4) ≥ (3n-1)/(5n²+4)
(3+1/n)/(5n+4/n) ≥ (3n-sin n)/(5n²+4) ≥ (3-1/n)/(5n+4/n)
lim n tend vers +infini (3+1/n)/(5n+4/n) = 0
lim n tend vers +nfini (3-1/n)/(5n+4/n) = 0
donc d apres le theoreme des gendarmes
lim n tend +infini (3n-sin n)/(5n²+4) = 0
voila j espere t avoir aider
