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lim[(x-tanx)/x^2sinx](x趋向于0)用洛必达法则
人气:372 ℃ 时间:2020-03-24 10:08:25
解答
lim(x→0)[(x-tanx)/x^2sinx] (用等价无穷小代换)
=lim(x→0)(x-tanx)/x^3 (0/0,用洛必达法则)
=lim(x→0)(1-sec^2x)/(3x^2)
=lim(x→0)-tan^2x/(3x^2) (用等价无穷小代换)
=lim(x→0)-x^2/(3x^2)
=-1/3答案是-1/2....答案肯定错误了。
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