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Math Help - Continuity proof

  1. #1
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    Continuity proof

    I need to show that x^2 is continuous on its domain using the epsilon-delta definition. So far, I've started with |(x^2) - (c^2)|< E, which is the same as |x+c||x-c|< E. However, I'm not quite how to get from here to |x-c|< D, where D is delta (which I'm pretty sure is supposed to depend on epsilon and c).
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  2. #2
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    Quote Originally Posted by spoon737 View Post
    I need to show that x^2 is continuous on its domain using the epsilon-delta definition. So far, I've started with |(x^2) - (c^2)|< E, which is the same as |x+c||x-c|< E. However, I'm not quite how to get from here to |x-c|< D, where D is delta (which I'm pretty sure is supposed to depend on epsilon and c).
    Let x_0 \in \mathbb{R} we want to show f(x) = x^2 is continous at that point. We want to show |f(x) - f(x_0)|< \epsilon can be made small, thus, |x^2 - x_0^2| = |x - x_0||x+x_0|< \epsilon. Now say |x-x_0|<\delta then |x| = |x-x_0+x_0|\leq |x-x_0| + |x_0| \implies |x|+|x_0| \leq |x-x_0|+2|x_0| but then |x+x_0|\leq |x|+|x_0|\leq |x-x_0|+2|x_0| < \delta + 2|x_0|. Thus, |f(x)-f(x_0)| = |x-x_0||x+x_0| < \delta (\delta + 2|x_0|) choose \delta \leq 1 then |\delta (\delta + 2|x_0|) \leq \delta (1+2|x_0|). Thus, we have that |f(x)-f(x_0|< \delta (1+2|x_0|), this means we have to chose \delta = \min \left\{ \frac{\epsilon}{1+2|x_0|}, 1\right\}
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  3. #3
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    Quote Originally Posted by kalagota View Post

    so, if \varepsilon > 0, choose \delta = \frac{\varepsilon}{1+2c}
    Careful what if c=-\frac{1}{2}
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Careful what if c=-\frac{1}{2}

    yeah, i should have edited it before i posted it.. anyways, you already had it.. thanks..

    hey, can you help me with my problem i posted in the algeb section? thanks.
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