求极限 x→∞lim{x²[1-xsin(1/x)]}令1/x=u,则x=1/u,x→∞时u→0.故原式=u→0lim{[1-(1/u)sinu]/u²}=u→0lim[(u-sinu)/u³]=u→0lim[(1-cosu)/(3u²)]=u→0lim[(sinu)/6u]=u→0lim[(cosu)/6]=1/6....