lim(x→0)[1/(sinx)^2-(cosx)^2/x^2]

我有人 1年前 已收到2个回答 举报

荒村听雨 春芽

共回答了22个问题采纳率:81.8% 举报

lim(x→0)[1/(sinx)^2-(cosx)^2/x^2]
=lim(x→0)[x^2-sin^2 x(cosx)^2]/[x^2(sinx)^2]
=lim(x→0)[x^2-sin^2 x(cosx)^2]/[x^4]
=lim(x→0)[x^2-1/4sin^2 (2x)]/[x^4] (0/0)
=lim(x→0)[2x-sin (2x)cos(2x)]/[4x^3]
=lim(x→0)[2x-1/2sin (4x)]/[4x^3] (0/0)
=lim(x→0)[2-2cos (4x)]/[12x^2]
=lim(x→0)(4x)^2/[12x^2]
=4/3

1年前

9

morfengmei 幼苗

共回答了967个问题 举报

lim(x->0)1/[1/sinx^2-cosx^2/x^2
=lim(x->0)x^2sinx^2/(x^2-sinx^2cosx^2)
=lim(x->0)x^2sinx^2/(x^2-(1/4)(sin2x)^2
=lim(x->0)(2xsinx^2+x^2sin2x)/(2x-(1/2)sin4x)
=lim(x->0)(2sinx^2+4xsin2x+...

1年前

1
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