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Math Help - prove function is continuous

  1. #1
    Senior Member sfspitfire23's Avatar
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    prove function is continuous

    If f(x) = 5x - 6, prove that f is continuous in its domain.

    So,

    Pick any > 0. Take any sequence { xn } converging to c. Then there exists an integer N such that
    | xn - c | < /5

    then,
    |f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|<


    My question is, why do we have |xn - c | < /5? Why not |xn - c | < ? Where did the 5 come from?


    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    If f(x) = 5x - 6, prove that f is continuous in its domain.

    So,

    Pick any > 0. Take any sequence { xn } converging to c. Then there exists an integer N such that
    | xn - c | < /5

    then,
    |f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|<


    My question is, why do we have |xn - c | < /5? Why not |xn - c | < ? Where did the 5 come from?


    Thanks.
    To counteract the coefficient of the x term?
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  3. #3
    Senior Member sfspitfire23's Avatar
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    I guess my question is how do you choose delta in continuous proofs when you dont know what the actual limit is?
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  4. #4
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    Answer

    See, the definition of continuity. the delta basically depends on the choice of epsilon. Moreover this function is uniformly continuous.

    you have f(x)=5x+6. AS per your proof we have x_{n} \to c \ \Rightarrow f(x_n) \to  f(c). Here u have the delta value as epsilon/5.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    I guess my question is how do you choose delta in continuous proofs when you dont know what the actual limit is?
    You don't in theory. But, I think you know what the limit should be.

    It's like induction. You can't use it until you know the answer, but usually the answer is a) apparent or b) unseeable.
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  6. #6
    Senior Member sfspitfire23's Avatar
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    So....really.....I should perform the |f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|< calculation first and then see what i need to make |xn-c| less than to satisfy the |f(xn)-f(c)| calculation?

    This seems odd to me.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    So....really.....I should perform the |f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|< calculation first and then see what i need to make |xn-c| less than to satisfy the |f(xn)-f(c)| calculation?

    This seems odd to me.
    Yes.
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  8. #8
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    Answer

    See, the definition of continuity. the delta basically depends on the choice of epsilon. Moreover this function is uniformly continuous.

    you have f(x)=5x+6. AS per your proof we have x_{n} \to c \ \Rightarrow f(x_n) \to  f(c). Here u have the delta value as epsilon/5.
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