设函数f(x)满足条件f(x+y)=f(x)+f(y),且f(x)在x=0处连续,证明f(x)在所有的点x0处连续 设函数f(x)在x=0处可导，讨论函数|f(x)|在x=0处...
\u5df2\u77e5\u5b9a\u4e49\u5728\u5b9e\u6570\u96c6R\u4e0a\u7684\u51fd\u6570y=f\uff08x\uff09\u6ee1\u8db3\uff1af\uff08x+y\uff09=f\uff08x\uff09+f\uff08y\uff09\uff0c\u4e14f\uff08x\uff09\u4e0d\u6052\u7b49\u4e8e\u96f6\uff0c\u5219y=f\uff08x\uff09\u662f\uff08\u8bc1\u660e\uff1a\u5f53x=y=0\u65f6\uff0c\u5219\u6709f\uff080\uff09=f\uff080\uff09+f\uff080\uff09=2\uff080\uff09\uff0c\u6240\u4ee5f\uff080\uff09=0\uff0c\u6240\u4ee5\u4ee4y=-x\uff0c\u5219\u6709f\uff080\uff09=f\uff08x\uff09+f\uff08-x\uff09=0\uff0c\u5219\u6709f\uff08-x\uff09=-f\uff08x\uff09\uff0c\u53c8\u56e0\u4e3a\u51fd\u6570\u7684\u5b9a\u4e49\u57df\u4e3aR\uff0c\u6240\u4ee5\u7b26\u5408\u5947\u51fd\u6570\u5b9a\u4e49\uff0c\u5373\u51fd\u6570y=f\uff08x\uff09\u662f\u5947\u51fd\u6570\uff0e\u6545\u9009A\uff0e
1. \u82e5\u51fd\u6570f(x)\u5728x=0\u7684\u67d0\u4e2a\u90bb\u57df\u5185\u4e0d\u53d8\u53f7\uff0c
\u5373\u5728\u8fd9\u4e2a\u90bb\u57df\u5185f(x)\u22650\u6052\u6210\u7acb\uff0c\u6216f(x)\u22640\u6052\u6210\u7acb\uff0c\u5219\u5728\u8fd9\u4e2a\u90bb\u57df\u5185|f(x)|=\u00b1f(x)\uff0c
\u663e\u7136\uff0c\u51fd\u6570|f(x)|\u5728x=0\u5904\u53ef\u5bfc\u3002
2. \u82e5\u51fd\u6570f(x)\u5728x=0\u7684\u4efb\u610f\u90bb\u57df\u5185\u53d8\u53f7\uff0c
\u5728\u8fd9\u4e2a\u90bb\u57df\u5185\uff0c
\u4e0d\u59a8\u8bbex>0\uff0c f(x)>0\uff0c
\u6709|f(x)|=f(x) \uff0c\u8fd9\u65f6|f(0+)|\u2019=f\u2019(0+)\uff1b
x<0\uff0cf(x)<0\uff0c\u6709|f(x)|=-f(x)\uff0c \u8fd9\u65f6|f(0-)|\u2019=-f\u2019(0-)\u3002
\u7531\u51fd\u6570f(x)\u5728x=0\u5904\u53ef\u5bfc\uff0c\u77e5f\u2019(0+)=f\u2019(0-).
\u53c8\u7531\u5047\u8bbe\u77e5\uff0cf\u2019(0)\u22600\uff0c\u5373f\u2019(0+)=f\u2019(0-)\u22600\uff08\u4e0d\u7136\u7684\u8bdd\uff0cx=0\u662ff(x)\u7684\u9a7b\u70b9\uff0cf(x)\u5728\u8fd9\u70b9\u5c06\u6539\u53d8\u589e\u51cf\u6027\uff0c\u4e0ef\u2019(0+)=f\u2019(0-)\u77db\u76fe\uff09
\u6240\u4ee5\uff0c \u51fd\u6570|f(x)|\u5728x=0\u5904\u4e0d\u53ef\u5bfc\u3002
\u4eb2\uff0c\u4e3e\u4f8b\u5982\u4e0b\u3002
1. y=cosx\uff0cy=-x²\u3002
2. y=sinx\uff0cy=x.
极限lim[f(x0+Δx)(Δx→0)存在且等于f(x0)。
因为f(x)在x=0处连续，所以limf(x)(x→0)=f(0)
又因为f(x+y)=f(x)+f(y), f(0)=f(0)+f(0)=2f(0), 所以f(0)=0
所以f(x0+Δx)=f(x0)+f(Δx)
所以lim[f(x0+Δx)(Δx→0)=limf(Δx)+f(0)(Δx→0)=f(x0)
即证明了函数在任意一点x处存在极限且等于f(x0)
结论得证
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