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Math Help - Proving a function is continuous

  1. #1
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    Proving a function is continuous

    Problem: Prove using epilson delta definition that f(x)=kx is continuous for any real number k.

    I have the k \neq 0 case. I just had some problems with the k=0 case. Here's what I got.

    Let \epsilon > 0 and choose \delta=\epsilon, then |x-x_{0}| < \delta implies that |x-x_{0}| < \epsilon and so |k||x-x_{0}| < |k|\epsilon . Then |kx-kx_{0}| < |k|\epsilon and since k=0 and f(x)=kx=0 and f(x_{0})=kx_{0}=0 then |f(x)-f(x_{0})| < |k|\epsilon . Since \epsilon > 0, then \epsilon > |k|\epsilon. So |f(x)-f(x_{0})| < \epsilon.

    Thoughts? Corrections?
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  2. #2
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    Re: Proving a function is continuous

    Quote Originally Posted by JSB1917 View Post
    Problem: Prove using epilson delta definition that f(x)=kx is continuous for any real number k.

    I have the k \neq 0 case. I just had some problems with the k=0 case.

    Oh, come on. If k=0 then f(x)=0 everywhere.

    Thus |x-a|<\delta implies |0-0|<\epsilon everywhere.
    Thanks from topsquark
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  3. #3
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    Re: Proving a function is continuous

    Quote Originally Posted by Plato View Post
    Oh, come on. If k=0 then f(x)=0 everywhere.

    Thus |x-a|<\delta implies |0-0|<\epsilon everywhere.
    Lol, this is the problem with drilling too much rigor into people. They start losing their sense of intuition.
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