求大神教用洛必达法则求X趋向于0时求（x-sinx）/x·sinx·arctanx的极限
1个回答

∵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.

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