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Консультация № 180192
07.10.2010, 19:25
0.00 руб.
07.10.2010, 20:38
0
3
1
Здравствуйте,уважаемые эксперты!
Помогите, пожалуйста, решить пределы функций:
https://rfpro.ru/upload/3222
1) limx → 1 (sin (ex - 1 - 1))/ln x;
2) limx → 0 (√(x ∙ arctg 4x))/(2x + arcsin 5x);
3) limx → ∞ x2 ∙ (cos (5/x) - cos (8/x)).
Спасибо!!! )))))))
Помогите, пожалуйста, решить пределы функций:
https://rfpro.ru/upload/3222
1) limx → 1 (sin (ex - 1 - 1))/ln x;
2) limx → 0 (√(x ∙ arctg 4x))/(2x + arcsin 5x);
3) limx → ∞ x2 ∙ (cos (5/x) - cos (8/x)).
Спасибо!!! )))))))
Обсуждение
07.10.2010, 20:07
общий
Миронычев Виталий:
Предлагаю Вам, как вариант, набирать формулы в редакторе MS Word, а затем использовать кнопочную панель в окне вопроса. Либо выполнять рисунки меньшего объёма, чтобы они легко открывались в окне браузера. Формат png бычно имеет больший объём, чем формат jpg. Используйте также изменение размеров рисунка в редакторе изображений.
Предлагаю Вам, как вариант, набирать формулы в редакторе MS Word, а затем использовать кнопочную панель в окне вопроса. Либо выполнять рисунки меньшего объёма, чтобы они легко открывались в окне браузера. Формат png бычно имеет больший объём, чем формат jpg. Используйте также изменение размеров рисунка в редакторе изображений.
Об авторе:
Facta loquuntur.
Facta loquuntur.
07.10.2010, 20:21
общий
Гордиенко Андрей Владимирович:
По-моему, все остальные вопросы Миронычева Виталия просто дублируют этот, адрес картинки один и тот же
https://rfpro.ru/upload/3222
По-моему, все остальные вопросы Миронычева Виталия просто дублируют этот, адрес картинки один и тот же
https://rfpro.ru/upload/3222
07.10.2010, 23:29
общий
это ответ
Здравствуйте, Миронычев Виталий.
1. При x → 1 (sin (ex - 1 - 1))/ln x = [0/0]. Сделаем подстановку t = x – 1. Тогда при x → 1 имеем t → 0, и
(sin (et – 1))/ln (t + 1) ~ (sin t)/t → 1 (здесь мы воспользовались эквивалентностями бесконечно малых функций и первым замечательным пределом). Ответ: 1.
2. При x → 0 (√(x ∙ arctg 4x))/(2x + arcsin 5x) ~ √(x ∙ 4x)/(2x + 5x) = 2x/(7x) → 2/7 (здесь мы воспользовались эквивалентностями бесконечно малых функций). Ответ: 2/7.
3. При x → ∞ x2 ∙ (cos (5/x) - cos (8/x)) = [∞ ∙ 0]. Сделаем подстановку t = 1/x. Тогда при x → ∞ t → 0, и
1/t2 ∙ (cos 5t – cos 8t) ~ 1/t2 ∙ ((5t)2/2 + 1 – (8t)2/2 – 1) = 1/t2 ∙ (-39t2/2) → -19,5 (здесь мы воспользовались эквивалентностями бесконечно малых функций). Ответ: -19,5.
С уважением.
1. При x → 1 (sin (ex - 1 - 1))/ln x = [0/0]. Сделаем подстановку t = x – 1. Тогда при x → 1 имеем t → 0, и
(sin (et – 1))/ln (t + 1) ~ (sin t)/t → 1 (здесь мы воспользовались эквивалентностями бесконечно малых функций и первым замечательным пределом). Ответ: 1.
2. При x → 0 (√(x ∙ arctg 4x))/(2x + arcsin 5x) ~ √(x ∙ 4x)/(2x + 5x) = 2x/(7x) → 2/7 (здесь мы воспользовались эквивалентностями бесконечно малых функций). Ответ: 2/7.
3. При x → ∞ x2 ∙ (cos (5/x) - cos (8/x)) = [∞ ∙ 0]. Сделаем подстановку t = 1/x. Тогда при x → ∞ t → 0, и
1/t2 ∙ (cos 5t – cos 8t) ~ 1/t2 ∙ ((5t)2/2 + 1 – (8t)2/2 – 1) = 1/t2 ∙ (-39t2/2) → -19,5 (здесь мы воспользовались эквивалентностями бесконечно малых функций). Ответ: -19,5.
С уважением.
5
спасибо вам огромное за такой аргументированный ответ
Об авторе:
Facta loquuntur.
Facta loquuntur.
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