lim(sin(x/2))^1/((1+cosx)) x到π 求极限
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lim(sin(x/2))^1/((1+cosx))x到π求极限lim(sin(x/2))^1/((1+cosx))x到π求极限lim(sin(x/2))^1/((1+cosx))x到π求极限1的无穷
lim(sin(x/2))^1/((1+cosx)) x到π 求极限
lim(sin(x/2))^1/((1+cosx)) x到π 求极限
lim(sin(x/2))^1/((1+cosx)) x到π 求极限
1的无穷大次方类型,取对数
令y=(sin(x/2))^[1/(1+cosx)]
lny=[1/(1+cosx)]ln(sin(x/2))
注意到:ln(sin(x/2))=ln(1+sin(x/2)-1)与sin(x/2)-1等价
lim(lny)=lim[1/(1+cosx)]ln(sin(x/2))
=lim [sin(x/2)-1]/(1+cosx)
洛必达法则
=lim (1/2)cos(x/2)/(-sinx) x→π
再洛必达法则
=lim -(1/4)sin(x/2)/(-cosx) x→π
=-(1/4) / 1
=-1/4
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢.
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