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x→0+時1/x→+∞所以lim(x→0+)arctan(1/x)→limarctan(+∞)=π/2x→0-時1/x→-∞所以lim(x→0-)arctan(1/x)→limarctan(-∞)=-π/2因為lim(x→0+)arctan(1/x)≠lim(x→0-)arctan(1/x)所以函數在該點的極限不存在…
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