Введение в математический анализ в вопросах и задачах. Анчиков А.М - 47 стр.

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x
1
= 4; x
2
=
√
6 + 4 =
=
√
10 , x
3
=
q
6 +
√
10 , x
4
=
r
6 +
q
6 +
√
10 , ... , x
n
=
=
v
u
u
t
6 +
s
6 +
r
6 + ... +
q
6 +
√
10.
3
c.
c =
√
6 + c.
c
2
−c − 6 = 0, c
1,2
=
1
2
±
5
2
. c
1
= 3, c
2
= −2.
3, x
n
{x
n
} {y
n
},
a) lim
n→∞
(
x
n
/
y
n
) = 0; b) lim
n→∞
(
x
n
/
y
n
) = c, c = const 6= 0;
c) lim
n→∞
(
x
n
/
y
n
) = ∞; d) {
x
n
/
y
n
}
a) lim
n→∞
3
√
n
2
·sin(n!)
n+2
; b) lim
n→∞
[
1
n
3
+
3
2
n
3
+...+
(2n−1)
2
n
3
];
c) lim
n→∞
(
1
2
+
3
2
2
+
5
2
3
+ ... +
2n−1
2
n
); d) lim
n→∞
n!+(n+2)!
(n−1)!+(n+2)!
.
[ a) 0; b)
4
3
; c) 3; d) 1 ].
a) lim
n→∞
2n
2
−2n+1
√
n
4
+1−n
; b) lim
n→∞
(n+3)
3
+(n+4)
3
(n+3)
4
−(n+4)
4
;
c) lim
n→∞
3
√
n
2
−
√
n
2
+5
5
√
n
7
−
√
n+1
; d) lim
n→∞
1
2
+3
2
+...+(2n−1)
2
2
2
+4
2
+...+(2n)
2
;
     Ðåøåíèå. Çäåñü ïîñëåäîâàòåëüíîñòü çàäàíà ðåêóððåíòíûì
                                              √
ñîîòíîøåíèåì. Íàéäåì îáùèé ÷ëåí:rx1 = 4; x2 =   6+4 =
  √           q    √                 q    √
= 10 , x3 =     6 + 10 , x4 =     6 + 6 + 10 , ... , xn =
     v        s
     u                   r
     u                                      q         √
     t
=        6+       6+         6 + ... +          6+        10. Ïîñëåäîâàòåëüíîñòü ìî-
íîòîííî óáûâàåò è îãðàíè÷åíà ñíèçó ÷èñëîì 3 . Òîãäà ñóùå-
ñòâóåò ïðåäåë, êîòîðûé îáîçíà÷èì ÷åðåç c. Òîãäà èç ðåêóððåíò-
                                                   √
íîãî ñîîòíîøåíèÿ íàõîäèì ( â ïðåäåëå) c = 6 + c. Îòñþäà
c2 − c − 6 = 0, c1,2 = 12 ± 25 . c1 = 3, c2 = −2. Ïðåäåëîì äàííîé
ïîñëåäîâàòåëüíîñòè ÿâëÿåòñÿ ÷èñëî 3, òàê êàê âñå ÷ëåíû xn
ïîëîæèòåëüíû.
  Ã. Çàäà÷è è óïðàæíåíèÿ äëÿ ñàìîñòîÿòåëüíîé ðàáî-
òû.
   1. Ïðèâåäèòå ïðèìåðû ïîñëåäîâàòåëüíîñòåé {xn } è {yn },
äëÿ êîòîðûõ:
    lim (xn/yn ) = 0;
a) n→∞                                 lim (xn/yn ) = c, c = const 6= 0;
                                   b) n→∞
c) lim (xn/yn ) = ∞; d) {xn/yn } ðàñõîäèòñÿ, íî îãðàíè÷åíà.
     n→∞

     2. Íàéòè ïðåäåëû:
          √
          3 2
           n ·sin(n!)                                                      2            (2n−1)2
a) lim      n+2
                      ;                                   b) lim [ n13 + n3 3 + ... +      n3
                                                                                                ];
      n→∞                                                     n→∞
                                                                    n!+(n+2)!
c)    lim ( 1   +   3
                    22
                         +   5
                             23
                                  + ... +   2n−1
                                             2n
                                                 );       d) lim                .
     n→∞ 2                                                    n→∞ (n−1)!+(n+2)!
                  4
[ a) 0; b)        3
                    ;    c) 3; d) 1 ].
     3. Âû÷èñëèòü ïðåäåëû:
          2n 2
          √ −2n+1 ;                                           (n+3)3 +(n+4)3
a) lim                                                b) lim       4       4;
      n→∞   n4 +1−n                                       n→∞ (n+3) −(n+4)
          √
          3 2 √ 2
            n − n +5                                         12 +32 +...+(2n−1)2
c)    lim √5 7 √     ;                            d) lim        2   2         2 ;
     n→∞    n − n+1                                       n→∞ 2 +4 +...+(2n)


                                                 47

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