Асимптотические оценки. Мицик М.Ф. - 3 стр.

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, 
()
xf

( ) ( ) ( ) ( )
......
1100
++++= xaxaxaxf
nn
ψψψ

x
, (1)

...,2,1,0
=
n

( ) ( ) ( ) ( ) ( )( )
xOxaxaxaxf
nnn 11100
......
+
+++++= ψψψψ
, (2)

x
.
  ,    
  .    (3) 
 
()
xf
,     
.
 4.
( )
dt
t
e
xf
x
t
=
1

+∞
x
. (3)
  (3)
n
, :
( )
( )
dt
t
e
n
t
n
ttt
exf
x
n
t
x
n
t
+
+
++++=
1
1
1
32
!
!1
...
!2!11

1
)1(
limlim
12
1
1
1
1
=
++
=
xnxn
xn
x
xn
x
tn
x
exexn
ex
ex
dtet
, 
( )
( )
+
++++=
+132
1!1
...
!2!11
nn
x
x
O
x
n
xxx
exf
, 
x
.

( )
...
!
...
!2!11
132
+++++=
+n
x
x
n
x
x
x
exf
(4)
(4) 
( )
xf
( )
...
!
...
!2!1
132
+++++=
+
n
xxxx
x
en
x
e
x
e
x
e
xf
(5)
 (5) .
( )
( )
( )
.11lim
1
!
!1
lim
!
:
!1
lim
2
1
12
>=+=
+
=
+
+
+
+
+
n
xnx
nx
x
en
x
en
n
n
n
n
n
x
n
x
n
,  (5) 
[
)
+∞;1
,
,  (5) 
 (3).
        ,                              f (x )
      f ( x ) = a0ψ 0 ( x ) + a1ψ 1 ( x ) + ... + anψ n ( x ) + ...                         x → ∞,          (1)
                            n = 0,1, 2, ...
   f ( x ) = a0ψ 0 ( x ) + a1ψ 1 ( x ) + ... + a nψ n ( x ) + ... + O (ψ n+1 ( x )) ,                       (2)
  x → ∞.
                                                        ,
                                                            .                                         (3)
                                  f (x ) ,
              .
                     4.
                                                    x
                                                    et
                                        f ( x ) = ∫ dt              x → +∞ .                                (3)
                                                  1
                                                    t
                                                        (3)     n                   ,            :

                                  (n − 1)! 
                                                                      x
                t 1
                                                                                x
                                                 et
      f ( x) = e  + 2 + 3 + ... + n  + n! ∫ n+1 dt
                     1! 2!
                  t t  t           t 1     1
                                               t
                            x

                            ∫t
                                 − n−1 t
                                       e dt
                                                                  x −n−1e x
                   lim      1
                                                = lim                                     = 1,
                   x→∞          x −n−1e x         x→∞ − ( n + 1) x − n−2 e x + x − n−1e x



      f (x )e x =
                          1 1! 2!          (n − 1)! + O 1 
                           + 2 + 3 + ... +              n+1  ,                            x → ∞.
                          x x   x            xn        x 

                                f (x)e x =
                                                1 1! 2!          n!
                                                 + 2 + 3 + ... + n +1 + ...                                 (4)
                                                x x   x         x
(4)                                                                                          f (x )
                            e − x 1!e − x 2!e − x n !e − x
                   f (x ) =      + 2 + 3 + ... + n+1 + ...                                                  (5)
                             x      x       x      x
                      (5)                                                               .

      lim
          (n + 1)!e − x n !e − x
                       :                        = lim
                                                       x n+1 (n + 1)! 1
                                                                     = lim(n + 1) = ∞ > 1.
      n →∞         x n+ 2              x n +1     n →∞    x n+ 2 n !  x n→∞
                     ,           (5)                                                          [1;+∞) ,
                      ,                                                   (5)
            (3).