求大神教用洛必达法则求X趋向于0时求（x-sinx）⼀x·sinx·arctanx的极限

解：∵lim(x->0)(sinx/x)=1，lim(x->0)(arctanx/x)=1 (都可以应用罗比达法则求得)
∴lim(x->0)[(x-sinx)/(x*sinx*arctanx)]
=lim(x->0){[(x-sinx)/x^3]/[(sinx/x)*(arctanx/x)]}
={lim(x->0)[(x-sinx)/x^3]}/{[lim(x->0)(sinx/x)]*[lim(x->0)(arctanx/x)]}
={lim(x->0)[(x-sinx)/x^3]}/(1*1)
=lim(x->0)[(x-sinx)/x^3]
=lim(x->0)[(1-cosx)/(3x^2)] (0/0型极限，应用罗比达法则)
=lim(x->0)[sinx/(6x)] (0/0型极限，应用罗比达法则)
=(1/6)*lim(x->0)(sinx/x)
=(1/6)*1
=1/6。