sinax~ax,
√(1-cosx)=√2sinx/2~√2x/2,
——》原式=limx→0+ =ax/(√2x/2)=√2*a.根号下1-cosx=根号下2sinx^2x吗1-cosx=2sin^2(x/2)。下面一步是ax除根号2x比2吗?是的。
lim(x-0+)sinax/根号下1-cosx,利用等价无穷小求极限
lim(x-0+)sinax/根号下1-cosx,利用等价无穷小求极限
数学人气:831 ℃时间:2019-10-17 01:59:01
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