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Math Help - Continuity proof

  1. #1
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    Continuity proof

    Show that f(x) = x^2 + 2x is continuous at 3.

    I know that I need to show that if |x - 3| < \delta then |x^2 +2x -15| < \epsilon. But I am not sure what to do after this.
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  2. #2
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    |x^2+2x-15|=|(x-3)(x+5)|=|x-3||x+5| < \epsilon

    If we demand that |x-3|< 1, i.e. 2<x<4, then |x+5|<9.

    If we also demand that |x-3|<\frac{\epsilon}{9}, then we get

    |(x-3)(x+5)|=|x-3||x+5| < 9\cdot \frac{\epsilon}{9}=\epsilon


    If we let \delta be the smallest of the two numbers 1 and \frac{\epsilon}{9}, then we have |x^2+2x-15|=|(x-3)(x+5)| < \epsilon when 0<|x-3|<\delta
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  3. #3
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    To be precise it has to be proven that for any real number \varepsilon>0, a real number \delta >0 can be found such that the implication |x-3|<\delta \Rightarrow |f(x)-f(3)|<\varepsilon holds.

    So the epsilon-delta game begins with epsilon.

    |f(x)-f(3)|<\varepsilon \Rightarrow |x-3||x+5|<\varepsilon,
    |x-3||x-5|=|x-3||x-3+8|\leq |x-3|(|x-3|+|8|)=|x-3|^2+8|x-3|

    Now, for arbitrarily chosen \varepsilon look at the inequality 0<|x-3|^2+8|x-3|<\varepsilon. Treat the expression |x-3| as you would any other normal variable (think of it as \delta) and you have a quadratic inequality which holds when |x-3|\in \left\langle 0, -4 +4\sqrt{1+\frac{\varepsilon}{16}}\right\rangle.

    To put it all in one place you have:
    For arbitrarily selected \varepsilon>0, you can use any \delta \in \left\langle 0, -4 +4\sqrt{1+\frac{\varepsilon}{16}}\right\rangle and the inequality |x-3|<\delta would imply that the inequality |f(x)-f(3)|<\varepsilon holds. Thus you have proven the continuity at 3.

    What Mondreus used as an example is simply a special case of this story: were you to choose \varepsilon=9 then you should use \delta \in \langle 0, -4+4\sqrt{1+\frac{9}{16}}\rangle=\langle 0,1\rangle.
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  4. #4
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    This makes sense to me, but I am not sure how you come up with |x - 3| < 1. Why 1? Thanks.

    Quote Originally Posted by Mondreus View Post
    |x^2+2x-15|=|(x-3)(x+5)|=|x-3||x+5| < \epsilon

    If we demand that |x-3|< 1, i.e. 2<x<4, then |x+5|<9.

    If we also demand that |x-3|<\frac{\epsilon}{9}, then we get

    |(x-3)(x+5)|=|x-3||x+5| < 9\cdot \frac{\epsilon}{9}=\epsilon


    If we let \delta be the smallest of the two numbers 1 and \frac{\epsilon}{9}, then we have |x^2+2x-15|=|(x-3)(x+5)| < \epsilon when 0<|x-3|<\delta
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