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Math Help - A proof involving continuity?

  1. #1
    Member mfetch22's Avatar
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    A proof involving continuity?

    Okay, so heres the problem:

    Suppose f is the function satisfying |f(x)| \leq |x| for all x. Show that f is continuous at 0.

    Okay, this is my work so far:



    We need to show \lim_{x \to 0} f(x) = f(0), I started by figuring the value of f(0). To find that value, I proceeded with:

    Since:

    x = 0

    and:

    |f(x)| \leq |x|

    then:

    |f(0)| \leq |0| \;\; \Rightarrow \;\; |f(0)| \leq 0

    but:

    |a| \geq 0 \;\;\;\; for \;\; all \;\; a

    So we must have:

    |f(0)| \geq 0

    But we also have the restriction that:

    |f(0)| \leq 0

    Therefore it must be true that:

    |f(0)| = 0 \;\; \Rightarrow \;\; f(0) = 0

    So, now i just need to show that:

    0 < |x-0| < \delta \;\; \Rightarrow \;\; |f(x) - f(0)| < \epsilon

    but since f(0) = 0; we need only show:

    0 < |x| < \delta \;\; \Rightarrow \;\; |f(x)| < \epsilon

    This is where i'm stuck. I betting the answer is right under my nose, and is really obivous, and I'm just not seeing it. I appreciate any help that anybody can give.
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  2. #2
    MHF Contributor

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    Just let \varepsilon  = \delta
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  3. #3
    Member mfetch22's Avatar
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    Quote Originally Posted by Plato View Post
    Just let \varepsilon  = \delta
    I'm not questioning your answer, I was just wondering if you could show me what about:

    0 < |x-0| < \delta

    and letting:

    \delta = \epsilon

    Implies that |f(x) - f(0)| = |f(x)| < \epsilon

    ??

    I'm guessing the answer is probably blatantly obvious or something, but I'm new to these proofs so I'm still learning. Thanks
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Prove something stronger:

    If f(x) defined on (a,b), and suppose there exist L\in\mathbb{R} so to every two points x_1 and x_2 in (a,b),

    |f(x_1)-f(x_2)|\leq L|x_1-x_2|

    then f(x) continuous on (a,b)
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  5. #5
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    \left| {f(x) - f(0)} \right| = \left| {f(x) - 0} \right| = \left| {f(x)} \right| \leqslant \left| x \right| < \delta  = \varepsilon
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  6. #6
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    Quote Originally Posted by mfetch22 View Post
    I'm not questioning your answer, I was just wondering if you could show me what about:

    0 < |x-0| < \delta

    and letting:

    \delta = \epsilon

    Implies that |f(x) - f(0)| = |f(x)| < \epsilon

    ??

    I'm guessing the answer is probably blatantly obvious or something, but I'm new to these proofs so I'm still learning. Thanks
    |f(x)| \le |x| < \delta

    EDIT: Darn you Plato. Beaten again.
    Last edited by mr fantastic; August 10th 2010 at 06:50 PM.
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