Results 1 to 4 of 4

Math Help - epsilon-delta definition

  1. #1
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722

    epsilon-delta definition

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

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

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

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

    any help appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Deadstar View Post
    I have a question on a tutorial involving using the \epsilon - \delta definition to prove a function is continuous. I missed the lectures on this so im not sure about it...

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

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

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

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

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

    now continue
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    k...

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

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

    \delta = min{ 1, epsilon/ (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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Deadstar View Post
    k...

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

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

    \delta = min{ 1, epsilon/ (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:

    |f(y) - f(b)| = |\sqrt[3]{y + 3} - \sqrt[3]{b + 3}| \le  \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|  = |(y + 3) - (b + 3)| = |y - b|

    i think you can finish

    i guess you would also need to show that \left( (y + 3)^{\frac 23} + \sqrt[3]{(y + 3)(b + 3)} + (b + 3)^{\frac 23} \right) \ge 1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Epsilon delta definition of limit.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 20th 2011, 06:20 PM
  2. epsilon-delta definition of continuity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 5th 2011, 09:13 AM
  3. epsilon delta definition
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 16th 2010, 08:26 AM
  4. epsilon delta-definition
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: August 27th 2009, 05:29 AM
  5. epsilon-delta definition of continuity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 3rd 2008, 07:51 AM

Search Tags


/mathhelpforum @mathhelpforum