2个回答
2017-10-15
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lim(x→π/2)(tanx)^sin2x
=e^[lim(x→π/2)sin2xlntanx
=e^[lim(x→π/2)sin2x/(1/lntanx)]
=e^[lim(x→π/2)2cos2xtanx/sec^2x]
=e^[lim(x→π/2)2cos2xsinxcosx]
=e^[lim(x→π/2)cos2xsin2x]
=e^[lim(x→π/2)1/2sin4x]
=e^[1/2sin(4×π/2)]
=e^0
=1
=e^[lim(x→π/2)sin2xlntanx
=e^[lim(x→π/2)sin2x/(1/lntanx)]
=e^[lim(x→π/2)2cos2xtanx/sec^2x]
=e^[lim(x→π/2)2cos2xsinxcosx]
=e^[lim(x→π/2)cos2xsin2x]
=e^[lim(x→π/2)1/2sin4x]
=e^[1/2sin(4×π/2)]
=e^0
=1
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