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Skvrtsvs
@Skvrtsvs
July 2022
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dimanchyck2000
По правилу Лопиталя:
lim (x → 0): (tg2x)'/(sin5x)' = 2/(cos²2x*5cos5x)= 2/(5cos²2x*cos5x) = 2/(5*1*1) = 2/5 = 0,4
Ответ: 0,4
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Answers & Comments
lim (x → 0): (tg2x)'/(sin5x)' = 2/(cos²2x*5cos5x)= 2/(5cos²2x*cos5x) = 2/(5*1*1) = 2/5 = 0,4
Ответ: 0,4