Введение в математический анализ в вопросах и задачах. Анчиков А.М - 45 стр.

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lim
n→∞
x
n
= a, ∀ n, x
n
> b (x
n
< b).
a) a > b, b) a ≥ b? ( c) a < b, d) a ≤ b ?)
a) lim
n→∞
3n
2
2−n
2
; b) lim
n→∞
[(n + 1)
k
− n
k
] , 0 < k < 1;
c) lim
n→∞
n [
q
n(n − 2) −
√
n
2
− 3 ] ;
d) lim
n→∞
3
n
−2
n
3
n+1
+2
n
; e) lim
n→∞
1+a+a
2
+...+a
n
1+b+b
2
+...+b
n
, |a| < 1, |b| < 1;
f) lim
n→∞
(
√
2 ·
4
√
2 · ... ·
2
n
√
2) ; g) lim
n→∞
3
√
n
2
+2−5n
2
n−
√
n
4
−n+1
.
a)
∞
∞
.
n
2
, lim
n→∞
3n
2
2−n
2
=
lim
n→∞
3
2
n
2
−1
=
lim
n→∞
3
lim
n→∞
(
2
n
2
−1)
= −3.
b) (∞−∞).
n
k
0 < (n + 1)
k
−n
k
= n
k
[ (1 +
1
n
)
k
−
−1] < n
k
[(1 +
1
n
) − 1] =
1
n
1−k
.
1
n
1−k
→ 0,
(n + 1)
k
− n
k
→ 0.
c)
lim
n→∞
n[
q
n(n − 2) −
√
n
2
− 3] =
= lim
n→∞
n [
√
n(n−2)−
√
n
2
−3] [
√
n(n−2)+
√
n
2
−3]
[
√
n(n−2)+
√
n
2
−3]
= lim
n→∞
n (n
2
−2n−n
2
+3)
[
√
n(n−2)+
√
n
2
−3]
=
= lim
n→∞
n(3−2n)
n[
√
1−
2
n
+
q
1−
3
n
2
]
= lim
n→∞
3−2n
√
1−
2
n
+
q
1−
3
n
2
= ∞.
   4. Ïóñòü lim xn = a, ïðè÷åì äëÿ ∀ n, xn > b (xn < b).
            n→∞
Ñëåäóåò ëè îòñþäà, ÷òî a) a > b, b) a ≥ b? ( c) a < b, d) a ≤ b ?)
Ïðèâåäèòå ïðèìåðû.
   Â. Ïðèìåðû ðåøåíèÿ çàäà÷.
   Ïðèìåð 31. Âû÷èñëèòü ïðåäåëû:
         3n2
a) lim      b) lim [(n + 1)k − nk ] , 0 < k < 1;
            2;
    n→∞ 2−n    n→∞
          q           √
c) lim n [ n(n − 2) − n2 − 3 ] ;
    n→∞
          3n −2n                               2          n
d) lim    n+1 +2n; e) lim 1+a+a     2
                                      +...+a
                                            n , |a| < 1, |b| < 1;
    n→∞ 3                 n→∞ 1+b+b +...+b
          √ √               √
                            n
                                              √3 2      2
                                       lim n−n√+2−5n
     lim ( 2 · 4 2 · ... · 2 2) ; g) n→∞
f ) n→∞                                             4
                                                   n −n+1
                                                          .
   Ðåøåíèÿ.
   a) Ýòîò ïðåäåë ÿâëÿåòñÿ íåîïðåäåëåííîñòüþ òèïà ∞ ∞
                                                      . ×èñ-
                                                        3n2
                                 2
ëèòåëü è çíàìåíàòåëü ïîäåëèì íà n , òîãäà ïîëó÷èì lim 2−n  2 =
                                                                                    n→∞
                       lim 3
 lim 2 3    =      n→∞
                 lim ( 22 −1)
                                       = −3.
n→∞ n2 −1
                n→∞ n

   b) Ìû èìååì çäåñü íåîïðåäåëåííîñòü âèäà (∞ − ∞). Ïðåîá-
ðàçóåì, âûíîñÿ nk çà ñêîáêè: 0 < (n + 1)k − nk = nk [ (1 + n1 )k −
−1] < nk [(1 + n1 ) − 1] = n1−k
                             1              1
                                . Òàê êàê n1−k → 0, òî è ïîäàâíî
(n + 1)k − nk → 0.
     c) Èçáàâèìñÿ îò èððàöèîíàëüíîñòè â ÷èñëèòåëå, óìíîæàÿ íà
ñîïðÿæåííîå âûðàæåíèå:
         q              √
  lim
 n→∞
       n[    n(n − 2) −   n2 − 3] =
             √        √       √       √
          n [ n(n−2)− n2 −3] [ n(n−2)+ n2 −3]        n (n2 −2n−n2 +3)
= n→∞
    lim             √         √
                                2
                                                 lim √
                                              = n→∞            √
                                                                  2
                                                                      =
                       [   n(n−2)+ n −3]                                  [    n(n−2)+ n −3]
               n(3−2n)                              3−2n
= lim     √         q                  = lim √        q                 = ∞.
  n→∞ n[         2
              1− n +       1−    3
                                   ]     n→∞        2
                                                 1− n +       1−    3
                                n2                                 n2



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