B. Szemberg WS 2004/05 Zahlenfolgen
L¨osungen zu dieser Aufgaben werden weder korrigiert (geben Sie sie nicht ab) noch in den ¨Ubungen be- sprochen.
Bestimmen Sie die folgenden Grenzwerte:
n→∞lim 1 + n1
= 1
n→∞lim (1−n) =−∞
n→∞lim
n−1 3n = 13
n→∞lim
3n+1 n = 3
n→∞lim
n+1 2n = 12
n→∞lim
2n−1 n+1 = 2
n→∞lim
5n+1 7−9n =−59
n→∞lim
n2+1 n =∞
n→∞lim
n2 (n+1)2 = 1
n→∞lim
n+3 n2−2 = 0
n→∞lim
5n2+1 n−3n2 =−53
n→∞lim
n+1 2n3 = 0
n→∞lim
2n2−4
−n−5 =−∞
n→∞lim
3n2−7n+1 2−5n−6n2 =−12
n→∞lim
(2n−1)2 n2+3n−1 = 4
n→∞lim
3n3−4n2+7 2n3+5n = 32
n→∞lim
n3−n
−2n4+n−5 = 0
n→∞lim
6n7−n4+2n3−1
−n5+2 =−∞
n→∞lim
(n+2)3−(n−2)3 95n3+39n = 0
n→∞lim 2n−1
5n+7 − 1+2n2+5n33
= 0
n→∞lim
√
n2+2n+n n+2 = 2
n→∞lim
n
√3
n3+1+n = 12
1
2
n→∞lim
√3
n2+3n n+4 = 0
n→∞lim
√3
n4+3n+1
n−1 =∞
n→∞lim
3n2−√ n3+1 n2−2n+4 = 3
n→∞lim
4n+1+1 22n+1−1 = 2
n→∞lim
3n+1−7 9n+4 = 0
n→∞lim
(2n+1)·(2n+2−4)
4n+9 = 4
n→∞lim
√n2+ 2−n
= 0
n→∞lim
√n2+n−√ n
=∞
n→∞lim
√n+ 1−√
n+ 2
= 0
n→∞lim p
(n+ 2)·(n+ 8)−n
= 5
n→∞lim
√n √
n+ 2−√ n
= 1
n→∞lim n32 √
n3+ 1−√
n3−2
= 32
n→∞lim lnn1 =−∞
n→∞lim 1 + 3n1 n
=e13
