Теорема 36. Если вблизи ! выполняются
неравенства ' < f < |
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lim |
'(x) = |
lim (x) = a; |
x!! |
x!! |
то
lim f(x) = a:
x!!
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Доказательство. |
9 |
=? |
lim |
'(x) = a |
(' < f < вблизи !) |
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x ! |
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lim |
(x) = a |
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x!! |
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lim f(x) = a () x!!
(8(xn); xn 2 A n f!g; и xn ! ! : f(xn) ! a) :
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Фиксируем произвольную (xn); xn 2 A n f!g; |
и xn |
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(' < f < |
вблизи !) |
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< f x |
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опр.19 |
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N = N( ) |
2 N |
такое, что |
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U (!)) |
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( n > N : '(x |
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x ! |
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lim |
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Из выделенного синим цветом следует, по
определению Гейне, что lim f(x) = a:
x!!
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3.19.О неопределённостях.
Пусть f; g : A ! B; A Rk; B R и пусть ! конечная или бесконечно удалённая предельная точка множества A.
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Теорема 37. Если |
lim |
1 и |
x!! f(x) = |
lim |
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x!! g(x) = 1, то |
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lim (f(x) + g(x)) = 1:
x!!
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9 f(x) = 1 =
lim (f(x) + g(x)) = 1 =)
x!!
(8(xn); xn 2 A n f!g; и xn ! ! :
f(xn) + g(xn) ! 1) :
First Prev Next Last Go Back Full Screen Close Quit
Фиксируем произвольную (xn); xn 2 A n f!g; и xn ! !.
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f(x) = |
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(f(xn) + g(xn) |
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g(x) = |
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(g(xn) |
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Из выделенного синим цветом следует, по
определению Гейне, что lim (f(x) + g(x)) =
x!!
1:
First Prev Next Last Go Back Full Screen Close Quit
Теорема 38. Если lim f(x) = 1 и функция
x!!
g ограниченная вблизи !, то
lim (f(x) + g(x)) = 1:
x!!
First Prev Next Last Go Back Full Screen Close Quit
Доказательство.
lim |
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1 |
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=? |
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x |
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! f(x) = |
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lim |
(f(x) + g(x)) = |
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(g - ограниченная |
вблизи |
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f(xn!) + g(xn) |
(8(xn); xn 2 A n f!g; и xn |
!1: |
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! 1 |
First Prev Next Last Go Back Full Screen Close Quit
Фиксируем произвольную (xn); xn 2 A n f!g;
и xn |
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9 |
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(f x |
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lim f x |
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(g - ограниченная вблизи !) |
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и U (!) такие, что |
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(xn |
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опр.2 |
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> |
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( N = N( ) |
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N такое, что |
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n > N : xn |
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U (!)) > |
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(f(xn) |
! 1 |
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! 1 |
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(f(x |
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> |
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((g(xn)) |
- ограниченная) |
= |
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Из выделенного синим цветом следует, по
определению Гейне, что lim (f(x) + g(x)) =
x!!
1:
First Prev Next Last Go Back Full Screen Close Quit
