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(α и β бесконечно малые одного порядка при x → ω)
h1, h2 R и Uµ(ω) такие, что x A ∩ Uµ(ω) :
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
<
Фиксируем ε0 = |a2| > 0.
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•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Пример 73. Пусть
α(x) = x(2 + cos x) и β(x) = sin 2x.
Показать, что это - бесконечно малые одного порядка при x → 0.
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Решение. Очевидно, что α и β бесконечно малые при x → 0. Так как
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sin 2x 2 sin x cos x |
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то, в силу теоремы 49, α и β бесконечно малые одного порядка при x → 0.
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
3.22.1. Эквивалентные бесконечно малые.
Пусть ω конечная или бесконечно удалённая предельная точка множества A Rk и
α, β : A → B, A Rk, B R бесконечно малые при x → ω.
Определение 87. Если α(x) и β(x) есть бесконечно малые при x → ω и
lim α(x) = 1,
x→ω β(x)
то говорят, что бесконечно малые α(x) и β(x) эквивалентны при x → ω и пишут:
α(x) β(x) при x → ω.
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Теорема 50. Бесконечно малые α(x) и β(x) эквивалентны при x → ω тогда и только тогда, когда их разность есть бесконечно малая более высокого порядка, чем α(x) ( и β(x)) при x → ω.
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Доказательство.
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опр.87 |
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α(x) |
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•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Теорема 51. Пусть
α(x) α1(x) и β(x) β1(x) при x → ω.
Тогда, если существует
lim α(x),
x→ω β(x)
то существует и
lim α1(x),
x→ω β1(x)
и эти пределы равны.
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Доказательство. |
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= lim α1(x)·α(x)· β(x) = x→ω α(x) β(x) β1(x)
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
Следствие 51.1. При вычислении пределов вида
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α1(x) . . . αk(x) |
lim |
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где все αi и βj |
- бесконечно малые при |
x → ω, можно заменять каждый из множителей αi, βj на величину, эквивалентную этому множителю, не меняя величины предела.
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit