Results 1 to 3 of 3

Math Help - I think I am making this harder

  1. #1
    Newbie
    Joined
    May 2006
    Posts
    20

    Unhappy I think I am making this harder

    the definition of continutiy is stated in garbled forms. Identify the one that is equivalent for the definition of continuity and draw a sketch of what each of the others represents.

    a. for every epsilon > 0 and every delta > 0, |x - c| < delta implies |f(x) - f(c)| < epsilon.

    b. There is an epsilon >0 such that for every delta>0, |x - c| <delta implies
    |f(x) - f(c)| < epsilon.

    c. for some epsilon > 0, there is a delta > 0 such that |x - c| < delta implies |f(x) - f(c)| < epsilon.

    d. There is a delta >0 such that for every epsilon >0, |x - c| <delta implies
    |f(x) - f(c)| < epsilon.


    I chose d. I don't know for sure.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Susie38
    the definition of continutiy is stated in garbled forms. Identify the one that is equivalent for the definition of continuity and draw a sketch of what each of the others represents.

    a. for every epsilon > 0 and every delta > 0, |x - c| < delta implies |f(x) - f(c)| < epsilon.

    b. There is an epsilon >0 such that for every delta>0, |x - c| <delta implies
    |f(x) - f(c)| < epsilon.

    c. for some epsilon > 0, there is a delta > 0 such that |x - c| < delta implies |f(x) - f(c)| < epsilon.

    d. There is a delta >0 such that for every epsilon >0, |x - c| <delta implies
    |f(x) - f(c)| < epsilon.


    I chose d. I don't know for sure.
    THe definition for coutinouity of real function is that,
    \lim_{x\to c}f(x)=f(c)
    By definition we have,
    \forall \epsilon>0 there is a \delta>0 such as, |f(x)-f(c)|<\epsilon whenever |x-c|<\delta.

    The meaning of "whenever" is the same as this implies something else. Thus it is a saying,
    .... |x-c|<\delta implies |f(x)-f(c)|<\epsilon thus (d) is the correct choice.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2006
    Posts
    53
    How do I use a graph to represent the other 3 choices that were not the right answer?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Harder limits
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 5th 2010, 06:15 AM
  2. Harder Integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 9th 2009, 12:26 PM
  3. harder gradient
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 18th 2008, 02:49 AM
  4. a little harder
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 29th 2007, 10:02 PM
  5. Here's another (harder) one
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 14th 2007, 09:50 PM

Search Tags


/mathhelpforum @mathhelpforum