lim(x→0)(1-cosx)[x-ln(1+tanx)]/(sinx)^4
人气:229 ℃ 时间:2019-10-10 06:40:33
解答
=lim(x→0)x^2/2*[x-ln(1+tanx)]/[x^4]
=lim(x→0)[x-ln(1+tanx)]/[2x^2]
=lim(x→0)[1-secx^2/(1+tanx)]/(4x)
=lim(x→0)(1+tanx-secx^2)/[4(1+tanx)sinx]
=lim(x→0)(cosx^2+cosx*sinx-1)/[4(1+tanx)sinxcosx^2]
=lim(x→0)(cosx-sinx)/[4(1+tanx)cosx^2]
=1/4
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