x→0+时1/x→+∞所以lim(x→0+)arctan(1/x)→limarctan(+∞)=π/2x→0-时1/x→-∞所以lim(x→0-)arctan(1/x)→limarctan(-∞)=-π/2因为lim(x→0+)arctan(1/x)≠lim(x→0-)arctan(1/x)所以函数在该点的极限不存在...