考研数学二
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发布时间:2022-04-26 09:26
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热心网友
时间:2022-06-26 17:41
lim<x→0>ln(1+x²)/(secx-cosx)
=lim<x→0>ln(1+x²)/[(1/cosx)-cosx]
=lim<x→0>cosx·ln(1+x²)/(1-cos²x)
=lim<x→0>ln(1+x²)/sin²x
=lim<x→0>[2x/(1+x²)]/(2sinxcosx)
=lim<x→0>[2x/(1+x²)]/(sin2x)
=lim<x→0>2x/(sin2x)
=1
热心网友
时间:2022-06-26 17:41
limx→0 [ln(1+x^2)]/(secx-cosx)
=limx→0 [2x/(1+x^2)]/(secxtanx+sinx)
=limx→0 2x/[(1+x^2)(secxtanx+sinx)]
=limx→0 2/[(1+x^2)(secxtanx/x+sinx/x)]
=limx→0 2/[(1+x^2)(secx+1)]
=limx→0 2/[(1+x^2)(secx+1)]
=1