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alex1046
@alex1046
September 2021
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Решите пожалуйста предел
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Arroga
Используем свойство второго замечательного предела и получаем:
lim cos(x)^(ctg(2x)/sin(3x)) = lim (1+(cos(x)-1))^(ctg(2x)/sin(3x)) = lim (1+(cos(x)-1))^(ctg(2x)/sin(3x) * (cos(x)-1))
Так как:
lim (ctg(2x)/sin(3x) * (cos(x)-1)) = -1/12 при x -> 2*pi, то
lim cos(x)^(ctg(2x)/sin(3x)) = e ^(-1/12)
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Answers & Comments
lim cos(x)^(ctg(2x)/sin(3x)) = lim (1+(cos(x)-1))^(ctg(2x)/sin(3x)) = lim (1+(cos(x)-1))^(ctg(2x)/sin(3x) * (cos(x)-1))
Так как:
lim (ctg(2x)/sin(3x) * (cos(x)-1)) = -1/12 при x -> 2*pi, то
lim cos(x)^(ctg(2x)/sin(3x)) = e ^(-1/12)