Question

高等数学简单求极限问题。。在线急等

(cos x/cos 2x)^1/x^2 当x趋于0的极限

Answer 1

解：∵lim(x->0)[(cosx-cos(2x))/(x²cos(2x))]
=lim(x->0)[(1/cos(2x))*((cosx-cos(2x))/x²)]
={lim(x->0)[1/cos(2x)]}*{lim(x->0)[(cosx-cos(2x))/x²]}
=1*{lim(x->0)[(cosx-cos(2x))/x²]}
=lim(x->0)[(2sin(3x/2)sin(x/2))/x²] (应用余弦差角公式)
=lim(x->0)[(3/2)(sin(3x/2)/(3x/2))(sin(x/2)/(x/2))]
=(3/2)*{lim(x->0)[sin(3x/2)/(3x/2)]}*{lim(x->0)[sin(x/2)/(x/2)]}
=(3/2)*1*1 (应用重要极限lim(z->0)(sinz/z)=1)
=3/2
∴lim(x->0)[(cosx/cos(2x))^(1/x²)]
=lim(x->0){[(1+(cosx-cos(2x))/cos(2x))^(cos(2x)/(cosx-cos(2x)))]^[(cosx-cos(2x))/(x²cos(2x))]}
={lim(x->0)[(1+(cosx-cos(2x))/cos(2x))^(cos(2x)/(cosx-cos(2x)))]}^{lim(x->0))[(cosx-cos(2x))/(x²cos(2x))]}
=e^{lim(x->0)[(cosx-cos(2x))/(x²cos(2x))]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^(3/2)。

追问

能看看我其他两个问题不？谢谢

追答

可以，请把链接发给我。

追问

http://zhidao.baidu.com/question/485970001.html?quesup2
http://zhidao.baidu.com/question/485955831.html?quesup2

Answer 2

取对数 y=(cos x/cos 2x)^1/x^2
limlny=[ln(cosx)-ln(cosx2x)]/x^2再使用若比特法则
lim[ln(cosx)-ln(cosx2x)]/x^2 =lim [-sin(x)/cosx+2sin2x/cos2x]/2x
=lim [-sin(x)cos2x+2sin2xcosx]/[(2x)cosxcos2x]
=lim [-(sinx/x)cos2x+4(sin2x/2x)cosx]/[2cosxxos2x]
=[-1×1+4×1×1]/[2×1×1]=3/2
limy=exp(3/2)

Answer 3

解答，原式当x趋于0时函数部分趋于1而另一部分趋于无限大因此结果是趋于无限大。

Answer 4

guanshui