3.14 Prouvons que suite (un)n∈N définie par un= 1
n converge vers 0.
Soit ε >0. Choisissons n0 ∈N avec n0 > 1
ε. Alors pour toutn>n0, on a que
|un−0|= 1 n −0
= 1 n
= 1 n 6 1
n0
< ε. On a ainsi montré que lim
n→+∞
1 n = 0.
1) lim
n→+∞
3n+ 1
n = lim
n→+∞
n(3 + 1n)
n = lim
n→+∞3 + 1n = lim
n→+∞3 + lim
n→+∞
1
n = 3 + 0 = 3 2) lim
n→+∞
2n−3
7n = lim
n→+∞
n(2−n3)
7n = lim
n→+∞
2−n3
7 = lim
n→+∞
1
7(2−n3) =
1 7 lim
n→+∞2−n3 = 17( lim
n→+∞2− lim
n→+∞
3
n) = 17(2−3· lim
n→+∞
1
n) = 17(2−3·0) = 27 3) lim
n→+∞
2n+ 3
n+ 1 = lim
n→+∞
n(2 + 3n)
n(1 + 1n) = lim
n→+∞
2 + 3n
1 + 1n
= lim
n→+∞2 + n3
n→+∞lim 1 + n1
= lim
n→+∞2 + 3· lim
n→+∞
1 n
n→+∞lim 1 + lim
n→+∞
1 n
= 2 + 3·0 1 + 0 = 2
1 = 2
4) lim
n→+∞
1
n2+n = lim
n→+∞
1
n2(1 + 1n) =
lim
n→+∞1
n→+∞lim
n2(1 + n1) = 1
( lim
n→+∞
n2)·( lim
n→+∞1 + 1n) = 1
( lim
n→+∞
n2)·( lim
n→+∞1 + lim
n→+∞
1 n) = 1
( lim
n→+∞
n2)·(1 + 0) = 1
n→+∞lim
n2 = lim
n→+∞
1
n2 = lim
n→+∞
1 n · 1
n = lim
n→+∞
1 n · lim
n→+∞
1
n = 0·0 = 0 5) lim
n→+∞
n2−3n
3n2+ 4 = lim
n→+∞
n2(1− n3)
n2(3 + n42) = lim
n→+∞
1− n3
3 + n42
= lim
n→+∞1− n3
lim
n→+∞3 + n42
= lim
n→+∞1− lim
n→+∞
3 n
n→+∞lim 3 + lim
n→+∞
4 n2
=
1−3· lim
n→+∞
1 n
3 + 4· lim
n→+∞
1 n2
= 1−3·0 3 + 4· lim
n→+∞
1 n · 1n
= 1
3 + 4· lim
n→+∞
1 n · lim
n→+∞
1 n
= 1
3 + 4·0·0 = 1 3
Analyse : limite et convergence d’une suite Corrigé 3.14
6) lim
n→+∞
−3n+ 2
n2+ 1 = lim
n→+∞
n(−3 + 2n)
n2(1 + n12) = lim
n→+∞
−3 + n2 n(1 + n12) = lim
n→+∞−3 +n2
n→+∞lim
n(1 + n12) =
lim
n→+∞−3 + lim
n→+∞
2 n
( lim
n→+∞
n)·( lim
n→+∞1 + n12) =
−3 + 2· lim
n→+∞
1 n
( lim
n→+∞
n)·( lim
n→+∞1 + lim
n→+∞
1 n2)
= −3 + 2·0 ( lim
n→+∞
n)·(1 + lim
n→+∞
1
n ·n1) = −3
( lim
n→+∞
n)·(1 + lim
n→+∞
1 n· lim
n→+∞
1 n) =
−3 ( lim
n→+∞
n)·(1 + 0·0) = −3
n→+∞lim
n = lim
n→+∞
−3
n =−3· lim
n→+∞
1
n =−3·0 = 0
Analyse : limite et convergence d’une suite Corrigé 3.14
