Zadania z matematyki
Granice ciągów
1. Korzystając z definicji granicy ciągu wykazać, że:
n + 2 1 2n2 + 3n + 1
1.1 lim = , 1.2 lim = 2,
n" n"
2n - 1 2 n2 + n + 1
n
1.3 (*) lim = 0, 1.4 lim (-1)n = 0.
8
n" n"
2n
2. Znalez
ć granice:
(n + 1)2 (n + 1)3 - (n - 1)3
2.1 lim , 2.2 lim ,
n" n"
2n2 (n + 1)2 + (n - 1)2
" " "
4 3
( n2 + 1 + n)2 n5 + 2 - n2 + 1
2.3 lim " , 2.4 lim " " ,
3 5 2
n" n"
n6 + 1 n4 + 2 - n3 + 1
n! 1
2.5 lim , 2.6 lim (1 + 2 + . . . + n),
n" n"
(n + 1)! - n! n2
( )
1 - 2 + 3 - 4 + . . . - 2n
2.7 lim " ,
n"
n2 + 1
( )
1 1 1
2.8 lim + + . . . + ,
n"
1 · 2 2 · 3 (n - 1) · n
1
2n - 1 2n - 1
2.9 lim , 2.10 lim ,
1
n" n"
2n + 1
2n + 1
" "
n n
2.11 lim 2n + 3n, 2.12 lim 3n - 2n,
n" n"
"
n100
n
2.13 lim n + 2n, 2.14 lim ,
n" n"
2n
" "
n n
2.15 lim 10n + 9n + 7n, 2.16 lim 5n - 3n + 2n,
n" n"
" "
n n
2.16 lim 3n4 + 2n2 + 1, 2.17 lim 2n3 - 3n2 + 15,
n" n"
1
( )
1 1 1
" "
2.18 lim " + + . . . + ,
n"
n2 + 1 n2 + 2 n2 + n
"
"
1 1
n
n2
2.19 lim 1 + + . . . + , 2.20 lim (n + 1)(n + 2)...2n,
n" n"
2 n
sin n n sin n!
2.21 lim lim , 2.22 lim ,
n"
x1
n (n + 1)
" "
"
" "
2.23 lim n( n + 3 - n), 2.24 lim n( n2 + 1 - n2 - 1),
n" n"
" "
"
1 + 2n2 - 4n2 - 1
3
2.25 lim , 2.26 lim n3 + 4n2 - n.
n" n"
n
3. Wyznaczyć granice ciągów:
( )n ( )2n-1
2n + 1 3n - 1
3.1 xn = , 3.2 xn = ,
2n - 2 3n + 1
( )2n+3 ( )n2
1 1
3.3 xn = 1 + )n , 3.4 xn = 1 + ,
n2 n
( )2n+3 ( )3n-1
n2 + 7n + 3 n2 - 1
3.5 xn = , 3.4 xn = ,
n2 + 2n - 1 n2 + 2n + 3
"
2
( )n ( )n +1
n n n2 + 3n + 1
3.6 xn = " , 3.7 xn = ,
3
n2 + 5n + 1
n3 + 2n2
( )2n+1 ( )n2
1 Ä„
3.8 xn = 1 + sin , 3.8 xn = cos .
n n
2