limx--0:(sinx)^2-x^2(cosx)^2\x^2(sinx)^2

lvzhidao 1年前 已收到1个回答 举报

wowogod 春芽

共回答了16个问题采纳率:87.5% 举报

lim(x→0)[(sinx)^2-x^2(cosx)^2]/[x^2(sinx)^2]
=lim(x→0)[(sinx)^2-x^2(cosx)^2]/x^4 (0/0)
=lim(x→0)[2sinxcosx-2x(cosx)^2+2x^2sinxcosx]/(4x^3)
=lim(x→0)cosx[sinx-xcosx+x^2sinx]/(2x^3)
=lim(x→0)[sinx-xcosx+x^2sinx]/(2x^3) (0/0)
=lim(x→0)[cosx-cosx+xsinx+2xsinx+x^2cosx]/(6x^2)
=lim(x→0)[3xsinx+x^2cosx]/(6x^2)
=lim(x→0)[3sinx+xcosx]/(6x) (0/0)
=lim(x→0)[3cosx+cosx-xsinx]/6
=2/3

1年前

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