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Math Help - [SOLVED] A weird limit proof

  1. #1
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    [SOLVED] A weird limit proof

    Hi, how would you solve the following problem

    Prove: If f(x) is defined on  \Re and continuous at x=0, and if f(x_{1}+x_{2})=f(x_{1})+f(x_{2}) \forall x_{1},x_{2} \in \Re , then f(x) is continuous at all  x \in \Re

    Thanks in advance

    By the way, is there a theorem that clearly states that any real number can be expressed as the sum of two other real numbers? I'm guessing this is a big part of the proof.

    **EDIT**
    Corrected the problem
    Last edited by akolman; March 24th 2008 at 01:33 PM.
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  2. #2
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    Quote Originally Posted by akolman View Post
    Hi, how would you solve the following problem

    Prove: If f(x) is defined on  \Re and continuous at x=0, and if f(x)=f(x_{1})+f(x_{2}) \forall x_{1},x_{2} \in \Re , then f(x) is continuous at all  x \in \Re

    Thanks in advance

    By the way, is there a theorem that clearly states that any real number can be expressed as the sum of two other real numbers? I'm guessing this is a big part of the proof.
    Do you mean to say f(x_1+x_2) = f(x_1)+f(x_2)?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Do you mean to say f(x_1+x_2) = f(x_1)+f(x_2)?
    Yes. I think I now get the problem, this changes everything.
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  4. #4
    Super Member PaulRS's Avatar
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    f(x+h)=f(x)+f(h)

    Taking the limit: \lim_{h\rightarrow{0}}f(x+h)=f(x)+\lim_{h\rightarr  ow{0}}f(h) (1)

    f(0+0)=f(0)+f(0) thus f(0)=0

    And now using the continuity around x=0 in (1) we get our result
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