Results 1 to 4 of 4

Math Help - Prove Function is Continuous using epsilon/delta

  1. #1
    Member
    Joined
    Jan 2007
    Posts
    114

    Prove Function is Continuous using epsilon/delta

    Here is the theorem i'm supposed to use to prove the continuity of the function f(x) = x^3.

    Let f be a real-valued function whose domain is a subset of R. Then f is continuous at x0 element dom(f) iff for every epsilon > 0 there exists delta > 0 s. t. x element dom(f) and |x - x0| < delta => |f(x)-f(x0)| < epsilon.

    So I don't know how to get delta. I'm pretty sure you are supposed to use the triangle inequality somewhere, and you need to work backwards but i'm just a bit confused.
    Thanks
    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 tbyou87 View Post
    Here is the theorem i'm supposed to use to prove the continuity of the function f(x) = x^3.

    Let f be a real-valued function whose domain is a subset of R. Then f is continuous at x0 element dom(f) iff for every epsilon > 0 there exists delta > 0 s. t. x element dom(f) and |x - x0| < delta => |f(x)-f(x0)| < epsilon.

    So I don't know how to get delta. I'm pretty sure you are supposed to use the triangle inequality somewhere, and you need to work backwards but i'm just a bit confused.
    Thanks
    I will do the hard part.

    Let \epsilon > 0. Choose some \delta > 0 to be determined, such that |x - x_0| < \delta when x \in dom(f). Furthermore, we may need an upper bound for |x + x_0| given that |x - x_0|< \delta so let us say |x - x_0|<1 \implies |x|<|x_0| + 1. So , by the triangle inequality, |x + x_0| \le |x| + |x_0| < 2 |x_0|+1. Now consider |f(x) - f(x_0)|. We have:

    |x^3 - x_0^3| = \left|(x - x_0) \left( x^2 + xx_0 + x_0^2 \right) \right| \le \left|(x - x_0) \left( x^2 + 2xx_0 + x^2\right) \right| = |(x - x_0)(x + x_0)^2| < |x - x_0|(2|x_0| + 1)^2

    Now, what does \delta have to be to make that last part
    < \epsilon?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2007
    Posts
    114
    Ok so I think delta = min{1, epsilon/(2|x0|+1)^2}

    I still have a question about why you thought of using the upperbound for |x+x0| and why you assume delta=1 at the beginning and not another value?
    Also, is that the general formula for solving these continuity problems?
    Thanks
    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 tbyou87 View Post
    Ok so I think delta = min{1, epsilon/(2|x0|+1)^2}
    correct

    I still have a question about why you thought of using the upperbound for |x+x0|
    i cheated, well not really, it's like when you want to find delta, you work backwards, so too did i work backwards here. i simplified |f(x) - f(x_0)| and realized i would need a bound for that

    and why you assume delta=1 at the beginning and not another value?
    because i felt like it. choosing 1/2 would work as well, so would 1/3, 2/5 ...

    1 is just a nice number to use, and it's the number my professor always uses

    Also, is that the general formula for solving these continuity problems?
    Thanks
    yes, this is more or less the general approach to this problems. there are both simpler and harder types to prove where you would have to modify your technique somewhat
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Epsilon Delta continuous proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 21st 2010, 06:41 AM
  2. Epsilon-delta definition of a continuous function
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 15th 2010, 05:20 PM
  3. Cauchy definition (epsilon-delta) of continuous functions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 28th 2009, 04:11 AM
  4. Replies: 0
    Last Post: October 27th 2009, 07:06 AM
  5. Replies: 0
    Last Post: December 2nd 2008, 12:43 PM

Search Tags


/mathhelpforum @mathhelpforum