Пример 55. Показать, что
lim ax = 0 (a > 1).
x→−∞
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= lim a−t
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Замечание. Функция f(t) := at, a > 1, бесконечно большая при t → +∞ (см. пример 51). Следовательно, в силу теоремы 29, функция
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малой при t → +∞.
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3.17. Теоремы о пределе функций.
Пусть ϕ, f : A → B, A Rk, B R и пусть ω конечная или бесконечно удалённая предельная точка множества A.
Теорема 30. Пусть lim f(x) = a,
x→ω
Тогда
lim (f(x) + ϕ(x)) = a +
x→ω
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Доказательство.
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lim (f(x) + ϕ(x)) = a + b Гейне |
x→ω |
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( (xn), xn A \ {ω}, и xn → ω :
f(xn) + ϕ(xn) → a + b) .
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Фиксируем произвольную (xn), xn A \{ω}, и xn → ω.
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(f(xn) + ϕ(xn) → a + b) .
Из выделенного синим цветом следует, по
определению Гейне, что lim (f(x) + ϕ(x)) =
x→ω
a + b.
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Теорема 31. Пусть lim f(x) = a, lim ϕ(x) = b. |
Тогда |
x→ω |
x→ω |
lim f(x)ϕ(x) = ab.
x→ω
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Доказательство.
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lim f(x) = a |
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lim ϕ(x) = b |
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lim f(x)ϕ(x) = ab Гейне |
x→ω |
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( (xn), xn A \ {ω}, и xn → ω :
f(xn) · ϕ(xn) → a · b) .
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Фиксируем произвольную (xn), xn A \{ω}, и xn → ω.
(f(xn) |
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(f(xn) · ϕ(xn) → a · b) .
Из выделенного синим цветом следует,
определению Гейне, что lim (f(x)ϕ(x))
x→ω
ab.
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Теорема 32. Пусть
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lim f(x) = a, lim ϕ(x) = b, |
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x→ω |
x→ω |
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причём b 6= 0. Тогда |
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x→ω ϕ(x) |
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Доказательство.
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