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数学
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求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
人气:197 ℃ 时间:2020-05-22 12:27:56
解答
x→0 lim (1-cos ax)/(sinx)^2
=x→0 lim (a*sin ax)/(2sinx*cosx)
=x→0 lim (a*sin ax)/sin2x
=x→0 lim (a^2*cos ax)/2cos2x
=(a^2*1)/(2*1)
=(a^2)/2
就是用两次洛必达法则就行了
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