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Thread: epsilon-delta definition

  1. #1
    Super Member Deadstar's Avatar
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    epsilon-delta definition

    I have a question on a tutorial involving using the $\displaystyle \epsilon - \delta$ definition to prove a function is continuous. I missed the lectures on this so im not sure about it...

    Let $\displaystyle f(y) = \sqrt[3] {y +3}$ and b > -3

    Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous at y = b.
    [hint use the identity $\displaystyle u^3 - v^3 = (u - v)(u^2 + uv + v^2)$]

    Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous from the right at y = -3

    any help appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Deadstar View Post
    I have a question on a tutorial involving using the $\displaystyle \epsilon - \delta$ definition to prove a function is continuous. I missed the lectures on this so im not sure about it...

    Let $\displaystyle f(y) = \sqrt[3] {y +3}$ and b > -3

    Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous at y = b.
    [hint use the identity $\displaystyle u^3 - v^3 = (u - v)(u^2 + uv + v^2)$]

    Prove, from the $\displaystyle \epsilon - \delta$ definition, that f(y) is continuous from the right at y = -3

    any help appreciated.
    You want to show that For all $\displaystyle \epsilon > 0$ there exists a $\displaystyle \delta > 0$ such that $\displaystyle y \in dom(f)$ and $\displaystyle |y - b|< \delta$ implies $\displaystyle |f(y) - f(b)| < \epsilon$

    here, use $\displaystyle u = \sqrt[3]{y + 3}$ and $\displaystyle v = \sqrt[3]{b + 3}$

    now continue
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  3. #3
    Super Member Deadstar's Avatar
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    k...

    Let $\displaystyle \epsilon > 0$, choose some $\displaystyle \delta > 0$ to be determines such that $\displaystyle |y - y_0| < \delta$ when $\displaystyle y \in dom(f) $. Upper bound for $\displaystyle |y + y_0|$ given that $\displaystyle |y - y_0| < \delta$ so let us say $\displaystyle |y - y_0| < 1 => |y| < |y_0 + 1|$. So by triangle inequality, $\displaystyle |y + y_0| \le |y| + |y_0| < 2|y_0| + 1$. Now consider $\displaystyle |f(y) - f(y_0)|$. We have:

    $\displaystyle |u^3 - v^3| = |(u - v)(u^2 + uv + v^2)| \le |(u - v)(u^2 + 2uv + v^2)| = $$\displaystyle |(u - v)|(u + v)^2 < |u - v|(2|y_0| + 1)^2$

    $\displaystyle \delta = min${ 1,$\displaystyle epsilon$/$\displaystyle (2|y_0| + 1)^2$}

    sorry but im not that great at the math equation thing yet so the last bit is a bit muddled. And somehow my font has gone red... weird...

    After this (if its right) i dont really know how to end it.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Deadstar View Post
    k...

    Let $\displaystyle \epsilon > 0$, choose some $\displaystyle \delta > 0$ to be determines such that $\displaystyle |y - y_0| < \delta$ when $\displaystyle y \in dom(f) $. Upper bound for $\displaystyle |y + y_0|$ given that $\displaystyle |y - y_0| < \delta$ so let us say $\displaystyle |y - y_0| < 1 => |y| < |y_0 + 1|$. So by triangle inequality, $\displaystyle |y + y_0| \le |y| + |y_0| < 2|y_0| + 1$. Now consider $\displaystyle |f(y) - f(y_0)|$. We have:

    $\displaystyle |u^3 - v^3| = |(u - v)(u^2 + uv + v^2)| \le |(u - v)(u^2 + 2uv + v^2)| = $$\displaystyle |(u - v)|(u + v)^2 < |u - v|(2|y_0| + 1)^2$

    $\displaystyle \delta = min${ 1,$\displaystyle epsilon$/$\displaystyle (2|y_0| + 1)^2$}

    sorry but im not that great at the math equation thing yet so the last bit is a bit muddled. And somehow my font has gone red... weird...

    After this (if its right) i dont really know how to end it.
    i don't think an upper bound like that is needed here. and you started off your list of inequalities wrong. |f(y) - f(b)| is not |u^3 - v^3|

    here's how you would begin:

    $\displaystyle |f(y) - f(b)| = |\sqrt[3]{y + 3} - \sqrt[3]{b + 3}| \le $ $\displaystyle \left| (\sqrt[3]{y + 3} - \sqrt[3]{b + 3}) \left( (y + 3)^{\frac 23} + \sqrt[3]{(y + 3)(b + 3)} + (b + 3)^{\frac 23} \right) \right| $ $\displaystyle = |(y + 3) - (b + 3)| = |y - b|$

    i think you can finish

    i guess you would also need to show that $\displaystyle \left( (y + 3)^{\frac 23} + \sqrt[3]{(y + 3)(b + 3)} + (b + 3)^{\frac 23} \right) \ge 1$
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