Granice ciągów  zadania
Zadanie 1
Obliczyć granicę ciągu (an ), gdy:
n 7n
2n2 -1
1 n + 1
ëÅ‚ öÅ‚ (o) an = n(n - n2 + 2) ëÅ‚ öÅ‚
(a) an =
ìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚ - 1 (v) an =
n2 + 1
2 n - 1Å‚Å‚
íÅ‚ Å‚Å‚ íÅ‚
(h) an =
2n+1
2n2
4n3 - n
1 (p) an = n(n - n2 + 2n)
ëÅ‚ öÅ‚
(b) an = ëÅ‚ öÅ‚
n2 + 3
ìÅ‚ ÷Å‚ - 2
ìÅ‚ ÷Å‚
(w) an =
n2 + 1
ìÅ‚ ÷Å‚
3
íÅ‚ Å‚Å‚
n2
íÅ‚ Å‚Å‚
n3 - n
3
n
(q) an = n - n3 - n2 n
(c) an =
(i) an = 5n7 - n2 + 2
n + 3
ëÅ‚ öÅ‚
3n5 + 1
(x) an =
ìÅ‚ ÷Å‚
4 - n
íÅ‚ Å‚Å‚
3
n
(r) an = n - n3 - n2
n
(j) an = 2n + 3n-1 + 2
ëÅ‚
1
ëÅ‚1 öÅ‚
ìÅ‚nöÅ‚
÷Å‚
(y) an = - ÷Å‚
ìÅ‚
íÅ‚2Å‚Å‚
(d) an = n2
íÅ‚ Å‚Å‚
n2 n - n2 + 2n
(k) an = 3n + n2 + 4
(s) an = 3n+1 Å" n!
(z) an =
2n+1 + 32n-1 n + 1
(e) an =
(n + 1)!3n
4n + 9n
(l) an = n - n2 + 4
n - n2 -1
2-n + 3 Å" 4n
(t) an =
(f) an =
(m) an = n - 2n2 -1
3n + 22n+1
2n - 4n2 + n
6n + 2n Å" 3n-1 + 5n
(g) an =
-
4n - 2 Å" 6n-1 (n) an = 2n - 4n2 n (u) an = sin n!
n + 1
Odpowiedzi:
(a) 2 1 1 (m) - " 1
(w) e6
(e) (h) (r)
(b) + " 1
(x) nie
3 2 3
(n)
(c) 0
istnieje
3 (i) 1 4 (s) 0
(f)
1
(y) 1
(j) 3 (o)  1 (t) 0
2
(d)
(z) 0
2 (u) 0
(k) + " (p) - "
(g)  4
(l) 0
(q) - "
(v) e14
Zadanie 2
Obliczyć granicę ciągu (an ), gdy:
5-2n
n2 +1 (e) an =1 + 2n - 3n
n + 4
ëÅ‚ öÅ‚
(a) an =
(j) an =
ìÅ‚ ÷Å‚
n+1
n
n + 3
íÅ‚ Å‚Å‚
2n + 1
ëÅ‚ öÅ‚
n
(f) an =
ìÅ‚ ÷Å‚
2n2 -3
n + 1
ëÅ‚ öÅ‚
n
íÅ‚ Å‚Å‚
ëÅ‚ -1÷Å‚
öÅ‚
n2
(b) an =
ìÅ‚ ÷Å‚
ìÅ‚
(k) an =
6n
2n
íÅ‚ Å‚Å‚
ìÅ‚ ÷Å‚
3n + 1
ëÅ‚ öÅ‚
n2
íÅ‚ Å‚Å‚
(g) an =
ìÅ‚ ÷Å‚
1 1 1
3n + 2
1 + + + K +
íÅ‚ Å‚Å‚
2 4 2n
n
(c) an =
n 3n + 2n
ëÅ‚ öÅ‚
1 + 3 + 5 + K + (2n -1) n
(h) an = (l) lim
ìÅ‚ ÷Å‚
n"
n + 1Å‚Å‚ 5n + 4n
íÅ‚
3n-1
n
1
ëÅ‚1 öÅ‚
(i) an = +
ëÅ‚ öÅ‚ ìÅ‚ ÷Å‚
n2 + 1÷Å‚1-n
(d) an = ìÅ‚
n
íÅ‚ Å‚Å‚
ìÅ‚ ÷Å‚
n
íÅ‚ Å‚Å‚
Odpowiedzi
(a) " (e) - " 3
(i) e3
(l)
(b) 0
(f) "
1
5
(j)
(c) 0
(g) e-2
e2
1
(k) e-2
(h)
(d) 0
e
Zadanie 3
KorzystajÄ…c z twierdzenia o trzech ciÄ…gach oblicz granice:
2n + (-1)n
(a) lim
n"
3n + 2
3n + 2n
n
(b) lim
n"
5n + 4n
2n2 + sin n!
(c) lim
n"
4n2 - 3cos n2
n
(d) lim 3 + sin n
n"