数学求极限：lim(x->0) (sinx-arctanx)⼀(x^2*In(1+x))=?

般情况下是不会等于零的，一旦你计算结果等于零很有可能就是错的，不排除是零的可能，但这种情况极少

lim(x->0) (sinx-arctanx)/(x^2*In(1+x))展开为二阶taylor
= lim(x->0) (x-1/6*x^3-(x-1/3*x^3))/(x^2* (x-1/2*x^2))
= lim(x->0) (1/6*x^3)/(x^3* (1-1/2*x))
= lim(x->0) (1/6)/(1-1/2*x)
=1/6

lim(x->0) (sinx-arctanx)/(x^2*In(1+x))

=lim(x->0) (sinx-arctanx)/(x^3) (0/0)
=lim(x->0) (cosx-1/√(1-x^2))/(3x^2)
=lim(x->0) (cosx*√(1-x^2)-1)/[3x^2*√(1-x^2)]
=lim(x->0) (cosx*√(1-x^2)-1)/(3x^2) (0/0)
=lim(x->0) (-sinx*√(1-x^2)-xcosx/√(1-x^2))/(6x)
=lim(x->0) [-sinx*(1-x^2)-xcosx]/[√(1-x^2)*(6x) ]
=lim(x->0) (-sinx+x^2*sinx-xcosx)/(6x) (0/0)
=lim(x->0) (-cosx+2x*sinx+x^2*cosx-cosx+xsinx)/6
=-1/3