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該極限不存在.【1】當x=(2n+1)π,(n=1,2,3,4,…)時,易知,恒有cosx=cos[2nπ+π]=-1.【2】當x=2nπ時,(n=1,2,3,.)易知,恒有cosx=(2nπ)=1.【3】若當x--->∞時,函數y=cosx的極限存在,易知,x沿著任何通路--->∞,函…
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