S EMINAR NOTES 4 (2. 11. 2011)
Compute the following limits.
1.
2n2 C n 3
;
n!1
n3 1
2n5 C 3n 2
;
n!1 n5
3n3 C 1
lim
2n3 C 6n
:
n!1 n3
7n C 7
lim
lim
2.
p
f nC1
3.
p
p p
p
1 C 2 C C n
n
ng; f. 1/ n. n C 1
n/g;
nC2
2
3
1 C 22 C C n2
1 C 23 C C n3
;
:
n3
n4
n
;
n!1 2n
lim
lim
4.
n!1
p
n
p
n
lim
n!1
5.
lim
!
an C b n
6.
lim . 1/n
a > b > 0:
for
a2n C b 2n
p
3
n3 C n
lim p
3
n!1
n3 C 2n
7.
2n
:
n!1 nŠ
p
n
2n C 4n ;
n!1
n
;
2
p
3
n3 C 1
:
p
3
n3 C n
p
n3 n n
p :
n3 C 2n n
.n C 4/100 .n C 3/100
:
n!1
.n C 2/100 n100
lim
8.
lim
n!1
9.
3n C n5
;
n!1 n6 C nŠ
lim
10.
p
n
n2 C n3 C n4 C 2n C 3n C 4n :
lim
p
3
n!1
p
3
nC1
p
4
n;
lim
n!1
p
lim .n2 C sin.n C 1//. n4 C 2
p
n!1
nC2
p
3
nC3
p
4
nC1
:
p
3
n
n4 C 1/:
11? .
3
5
2n 1
1
C 2 C 3 C C
;
lim
n!1 2
2
2
2n
1
1
1
1
lim
1
1
1
1
:
n!1
22
32
42
n2
12.
p
p
3
n2 C 7 3 n2 C 1
lim
;
p
p
3
3
n!1
n2 C 6
n2
r
13. Does limn!1 xn exist, where xn D
s
lim
n
n!1
..n C 2/2
..n C 1/3
q
p
2 C ::: 2 ?

2C
n square roots
1
.n C 1/2 /nC1
:
n3 3n2 /n 1
2
2. S OLUTIONS
1. 0, 2, 2
2. 0, the limit does not exist, 1=2, 1=3, 1=4
3. 0, 4, 0
4. 1=a
5. First we write
p
3
n3 C n
p
3
n3 C 2n
p
3
n3 C 1
p
3
n3 C n
1
D
1
.n3 C n/ 3
.n3 C 1/ 3
1
1
.n3 C 2n/ 3
.n3 C n/ 3
2
1
1
2
2
1
1
2
.n3 C n/ 3 C .n3 C n/ 3 .n3 C 1/ 3 C .n3 C 1/ 3
.n3 C n/ 3 C .n3 C n/ 3 .n3 C 1/ 3 C .n3 C 1/ 3
2
1
1
2
2
1
1
2
.n3 C 2n/ 3 C .n3 C 2n/ 3 .n3 C n/ 3 C .n3 C 1/ 3
(1)
.n3 C 2n/ 3 C .n3 C 2n/ 3 .n3 C n/ 3 C .n3 C 1/ 3
D
n
1
2
1
2
1 .n3 C 2n/ 3 C .n3 C 2n/ 3 .n3 C n/ 3 C .n3 C n/ 3
2
1
1
2
n
.n3 C n/ 3 C .n3 C n/ 3 .n3 C 1/ 3 C .n3 C 1/ 3
2
1
1
2
2
1
1
2
1 .1 C n2 / 3 C .1 C n2 / 3 .1 C n2 / 3 C .1 C n2 / 3
D
2
1
1
2
n
.1 C n12 / 3 C .1 C n12 / 3 .1 C n13 / 3 C .1 C n13 / 3
p
p
3
n3 C n 3 n3 C 1
D 1:
It follows lim p
p
3
n!1 3 n3 C 2n
n3 C n
6. We start with the following limit
p
n3 n n
lim
p :
n!1 n3 C 2n n
We have
p
p
n
n
n3 n n
lim 3
D
lim
p
q np D 1:
2n
n!1 n C
n!1
n
n
1C
n6
p
We used thorem on arithmetics of limit and limn!1 n n D 1. From above and from theorem on limit of
subsequence we get
p
p
.2n/3 2n 2n
.2n C 1/3 2nC1 2n C 1
lim
a
lim
D 1:
p D1
p
n!1 .2n/3 C 4n 2n
n!1 .2n C 1/3 C 4nC2 2n C 1
p
n3 n n
This gives that lim . 1/n 3
p does not exist.
n!1
n C 2n n
7. We have
!
!
100
100
X
100 100 j j X 100 100 j j
100
100
.n C 4/
.n C 3/
D
n
4
n
3
j
j
j D0
j D0
!
100
X
(2)
100 100 j j
D
n
.4
3j / D 100n99 C P1 .n/I
j
n
j D0
100
.n C 2/
100
n
D 200n99 C P2 .n/;
where P1 ; P2 are polynomials of degree less than 99. We have
P1 .n/
D 0;
n!1 n99
lim
P2 .n/
D 0:
n!1 n99
lim
3
We get
100 C
.n C 4/100 .n C 3/100
D lim
100
100
n!1
n!1
.n C 2/
n
200 C
lim
P1 .n/
n99
P2 .n/
n99
D 1=2:
2n
D C1. This implies that there exists n0 2 N such
n!1 nk
8. We know that for each k 2 N it holds lim
that
8n 2 N; n n0 W n2 n3 n4 2n 3n 4n :
Thus we have
p
p
n
n
8n 2 N; n n0 W 4 n2 C n3 C n4 C 2n C 3n C 4n 4 6:
p
n
We know that lim
6 D 1 and therefore according to sandwich theorem we get
n!1
p
n
n2 C n3 C n4 C 2n C 3n C 4n D 4:
lim
n!C1
9. 0; 0; 0
10. We have
p
lim .n2 C sin.n C 1// . n4 C 2
n!1
p
n4 C 1/
p
p
p
n4 C 2 C n4 C 1
n4 C 1/ p
p
n!C1
n4 C 2 C n4 C 1
1 C sin.nC1/
n2 C sin.n C 1/
n2
D lim p
:
D lim q
q
p
n!C1
n!C1
2
n4 C 2 C n4 C 1
1 C n4 C 1 C n14
p
D lim .n2 C sin.n C 1// . n4 C 2
We have
8n 2 N W j sin.n C 1/j 1,
limn!1 n12 D 0,
p
is continuous on its domain.
D 0. This, (3) and arithmetics of limit we get
From (1) and (2) it follows limn!1 sin.nC1/
n2
p
p
1
lim .n2 C sin.n C 1// n4 C 2
n4 C 1 D :
n!1
2
11. 3, 1=2
12. 1, 2=3
13. 2
(3)