MATEMATYKA - LISTA 2
14.10.2016
ZAD.1 Oblicz granice poni˙zszych ci ,
ag´ow:
an=n2−1
3−n3;bn=(n−1)(n+ 3)
3n2+ 5 ;cn=(2n−1)3
(4n−1)2(1 −5n);dn=5n−2
3n−13
;en=n−10
3;
fn=(−1)n
2n−1;gn=√n−2
3n+ 5 ;hn=2−5n−10n2
3n+ 15 ;in=√1+2n2−√4n2+ 1
n;jn=3
rn−1
8n+ 10
kn=n
3
√8n3−n−n;ln=1
√4n2+ 7n−2n;mn=√n+ 2 −√n;on=pn2+n−n;
pn=p3n2+ 2n−5−n√3; rn=n3
√2−
3
p2n3+ 5n2−7.
ZAD.2 Oblicz granice poni˙zszych ci ,
ag´ow:
an=4n−1−5
22n−7;bn=5·32n
−1
4·9n+ 7 ;cn=3
2n
·
2n+1 −1
3n+1 −1;dn=n
√3n+ 2n;en=n
√10n+ 9n+ 8n;
fn=n
√10100 −
n
r1
10100 .
ZAD.3 Oblicz granice poni˙zszych ci ,
ag´ow (skorzystaj z Tw. o trzech ci ,
agach):
an=n
√2·3n+ 4 ·7n;bn=n
p3n+ sin (n2016 + 45); cn=n
r(−1)n
n+ 2n;dn=[n√2]
[n√3];
en=n1
n2+ 1 +2
n2+ 2 +. . . +n
n2+n;fn=n
s2
3n
+3
4n
;gn=[√2n]
n;hn= logn+1 (n2+ 1).
ZAD.4 Oblicz granice poni˙zszych ci ,
ag´ow:
an=1 + 1
n+ 23n
;bn=1−
1
nn
;cn=1−
1
n22n+1
;dn=1 + 1
2n2n+1
;en=3n+ 1
3n+ 4n
;
fn=n−1
n+ 32n+1
;gn=n2+ 3n+ 1
n2+ 5n−13n2+4n−7
.
ZAD.5* Korzystaj ,
ac z twierdzenia o ci ,
agu monotonicznym i ograniczonym uzasadnij, ˙ze poni ˙zsze ci ,
agi
maj ,
a granic,
e sko´nczon ,
a. Je´sli si,
e da, to oblicz te granice.
an=100n
n!;bn=[(3n)!]2
(2n)!(4n)!;c1=√2, cn+1 =√2 + cn;d1= 1, dn+1 = 1 + 1
1 + dn
;
en=1
n+ 1 +1
n+ 2 +. . . +1
n+n;fn=1
1! +1
2! +. . . +1
n!;gn=n!
nn.
1