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AnGrey18
@AnGrey18
July 2022
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Кто шарит в тригонометрии, help me плз. lim[x=>0] ((cos2x-2cosx+cos4x)/x^2cosx)
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ShirokovP
Verified answer
Неопределенность 0/0
Правило Лопиталя
(cos2x - 2cosx + cos4x) ' = - 2sin2x + 2sinx - 4sin4x
(x^2 cosx)' = 2xcosx - x^2 sinx
lim x->0 ( -2*0 + 2*0 - 4*0)/(2*0 - 0)
Неопределенность 0/0
Правило Лопиталя
(- 2sin2x + 2sinx - 4sin4x)' = 2cosx - 4cos2x - 16cos4x
(2xcosx - x^2 sinx )' = - (x^2 - 2)cosx - 4xsinx
lim x->0 (2 - 4 - 16)/(2 - 0) = - 18/2 = - 9
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AnGrey18
спасибо большое ♡
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Answers & Comments
Verified answer
Неопределенность 0/0Правило Лопиталя
(cos2x - 2cosx + cos4x) ' = - 2sin2x + 2sinx - 4sin4x
(x^2 cosx)' = 2xcosx - x^2 sinx
lim x->0 ( -2*0 + 2*0 - 4*0)/(2*0 - 0)
Неопределенность 0/0
Правило Лопиталя
(- 2sin2x + 2sinx - 4sin4x)' = 2cosx - 4cos2x - 16cos4x
(2xcosx - x^2 sinx )' = - (x^2 - 2)cosx - 4xsinx
lim x->0 (2 - 4 - 16)/(2 - 0) = - 18/2 = - 9