问题标题:
用洛必塔法则求极限lim(x趋于0)(1/sinx^2)-(1/x^2)
问题描述:
用洛必塔法则求极限lim(x趋于0)(1/sinx^2)-(1/x^2)
高宇回答:
lim(x→0)(1/sinx^2)-(1/x^2)(通分)
=lim(x→0)(x^2-sinx^2)/(x^2sin^2x)(等价无穷小代换)
=lim(x→0)(x^2-sinx^2)/(x^4)(0/0,洛必达法则)
=lim(x→0)(2x-2sinxcosx)/(4x^3)
=lim(x→0)(x-1/2sin2x)/(2x^3)(0/0,洛必达法则)
=lim(x→0)(1-cos2x)/(6x^2)(等价无穷小代换)
=lim(x→0)1/2(2x)^2/(6x^2)
=2/6
=1/3
查看更多
八字精批
八字合婚
八字起名
八字财运
2024运势
测终身运
姓名详批
结婚吉日