Введение в математический анализ в вопросах и задачах. Анчиков А.М - 46 стр.

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d) lim
n→∞
3
n
−2
n
3
n+1
+2
n
= lim
n→∞
1−(2/3)
n
3+(2/3)
n
=
lim
n→∞
[1−(2/3)
n
]
lim
n→∞
[3+(2/3)
n
]
=
1
3
.
e) lim
n→∞
1+a+a
2
+...+a
n
1+b+b
2
+...+b
n
= lim
n→∞
(
1−a
n+1
1−a
·
1−b
1−b
n+1
) =
=
1−b
1−a
· lim
n→∞
1−a
n+1
1−b
n+1
=
1−b
1−a
·
lim
n→∞
(1−a
n+1
)
lim
n→∞
(1−b
n+1
)
=
1−b
1−a
;
f) lim
n→∞
(
√
2 ·
4
√
2 · ... ·
2
n
√
2) = lim
n→∞
2
1/2+1
/
2
2
+...+1/2
n
=
= lim
n→∞
2
1
2
·
1−(1/2)
n
1−(1/2)
= lim
n→∞
2
1−(1/2)
n
= 2 lim
n→∞
2
−1/2
n
= 2;
g) lim
n→∞
3
√
n
2
+2−5n
2
n−
√
n
4
−n+1
= lim
n→∞
(
3
√
n
2
+2
/
n
2
)−5
(1/n−
q
1−1
/
n
3
+1
/
n
4
)
=
= lim
n→∞
3
q
1
/
n
4
+2
/
n
6
−5
(1/n−
q
1−1
/
n
3
+1
/
n
4
)
= 5.
lim
n→∞
x
n
, x
n
=
3
√
n + 1 −
3
√
n − 1.
a
m
− b
m
= (a − b)(a
m−1
+ a
m−2
b + a
m−3
b
3
+ ... + ab
m−2
+ b
m−1
).
a =
3
√
n + 1, b =
3
√
n − 1, m = 3. lim
n→∞
(
3
√
n + 1−
−
3
√
n − 1) = lim
n→∞
(
3
√
n+1−
3
√
n−1) · (
3
√
(n+1)
2
+
3
√
n+1 ·
3
√
n−1 +
3
√
(n−1)
2
)
(
3
√
(n+1)
2
+
3
√
n+1
3
√
n−1 +
3
√
(n−1)
2
)
=
= lim
n→∞
n+1−n+1
(
3
√
(n+1)
2
+
3
√
n
2
−1 +
3
√
(n−1)
2
)
= 0,
∞.
{x
n
} x
1
= 4; x
n+1
=
√
6 + x
n
.
                     n     n                          n         lim [1−(2/3)n ]
    d) n→∞   3 −2
        lim 3n+1 +2n
                        lim 1−(2/3)
                     = n→∞  3+(2/3)n
                                     =                        n→∞
                                                                lim [3+(2/3)n ]
                                                                                      = 13 .
                                                              n→∞
                    2       n                         n+1
    e)     lim 1+a+a2 +...+an        =      lim ( 1−a            ·    1−b
                                                                     1−bn+1
                                                                            )     =
          n→∞ 1+b+b +...+b                 n→∞     1−a

                         n+1
                                            lim (1−an+1 )
    1−b
=   1−a
             lim 1−a
          · n→∞  1−bn+1
                        =       1−b
                                1−a
                                       ·   n→∞
                                            lim (1−bn+1 )
                                                                 =    1−b
                                                                      1−a
                                                                          ;
                                           n→∞
             √ √               √
    f ) lim ( 2 · 4 2 · ... · 2 2) = lim 21/2+1/2 +...+1/2 =
                               n                 2        n

          n→∞                                             n→∞
                       n
              1 1−(1/2)
               ·                                  n                           n
= lim 2       2 1−(1/2)    = lim 21−(1/2) = 2 lim 2−1/2 = 2;
    n→∞                        n→∞                            n→∞
               √                                        √
                3 2       2                           ( 3 n2 +2/n2 )−5
    g)     lim n−n√+2−5n
                    n4 −n+1
                                 = lim                    q                       =
          n→∞                         n→∞ (1/n−             1−1/n3 +1/n4 )
                q
                3
                    1/n4 +2/n6 −5
= lim               q                      = 5.
    n→∞ (1/n−         1−1/n3 +1/n4 )
                                                                              √                √
    Ïðèìåð 32. Íàéòè n→∞
                      lim xn , åñëè xn =                                      3
                                                                                  n+1−         3
                                                                                                   n − 1.
    Ðåøåíèå. Ïðè âû÷èñëåíèè áóäåì ïîëüçîâàòüñÿ ôîðìóëîé
am − bm = (a − b)(am−1 + am−2 b + am−3 b3 + ... + abm−2 + bm−1 ).
                       √                √                          √
 íàøåì ñëó÷àå a = 3 n + 1, b = 3 n − 1, m = 3. lim ( 3 n + 1−
                                                            n→∞
 √                 √      √          √          √       √        √
                 ( 3 n+1− 3 n−1) · ( 3 (n+1)2 + 3 n+1 · 3 n−1 + 3 (n−1)2 )
− 3 n − 1) = lim            √
                            3           √
                                        3     √
                                              3
                                                       √
                                                       3
                                                                           =
                     n→∞             2(      (n+1) +        n+12      n−1 +       (n−1) )

   lim √
= n→∞                   n+1−n+1
                         √             √                  = 0, òàê êàê çíàìåíàòåëü ñòðå-
       3
          (     (n+1)2 + 3 n2 −1 +     3
                                            (n−1)2 )

ìèòñÿ ê ∞.
    Ïðèìåð 33. Íàïèñàòü îáùèé ÷ëåí, äîêàçàòü ñõîäèìîñòü ïî-
ñëåäîâàòåëüíîñòè {xn } è íàéòè å¼ ïðåäåë, åñëè x1 = 4; xn+1 =
√
  6 + xn .



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