Question

求极限lim(x趋向于0) (x-tanx)/xsinx^2

Answer 1

lim(x-tanx)/xsinx^2
=lim(x-tanx)/(x * x^2 * sinx^2 / x^2) 等价无穷小量:
=lim(x-tanx)/(x * x^2)
=lim(1/x^2 - sinx/x * 1/[x^2 * (cosx)]
=lim(1 - secx)/x^2 罗必塔法则：
=lim(-tanxsecx)/2x 罗必塔法则：
=-1/2limsecx(tan^2 x + sec^2 x)
=-1/2

追问

我觉得你第三个等号时算错了，请修改...

追答

没算错啊，lim(x-tanx)/(x * x^2)=lim1/x^2 - sinx/x * 1/（x^2 * cosx）把tanx=sinx/cosx

追问

我只算到(xcosx-sinx)/x^2*cosx,算不到你的第三个=号那条式，请讲解一下

追答

(x-tanx)/(x * x^2)
=x/(x * x^2) - tanx/(x * x^2)
=1/x^2 - (sinx/cosx)/(x * x^2)
=1/x^2 - sinx/x * 1/(cosx * x^2)