Results 1 to 5 of 5

Math Help - Precise Definition of A Limit Problem

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    12

    Precise Definition of A Limit Problem

    Stuck on this one problem in my math homework:
    http://i.imgur.com/2uVLo.png
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Nov 2011
    Posts
    12

    Re: Precise Definition of A Limit Problem

    I know how to find the precise definition of a limit but the way this one is set up is weird to me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Precise Definition of A Limit Problem

    we want to find a δ > 0 so that when:

    |x - 1/2| < δ, then

    |f(x) - L| < ε = c

    now f(x) = mx + b, and L = m/2 + b, so we need to find such a δ that makes:

    |mx + b - (m/2 + b)| < c.

    note that:

    |mx + b - (m/2 + b)| = |mx + b - m/2 - b| = |m(x - 1/2)| = |m||x - 1/2| = m|x - 1/2| (since m > 0).

    so what we really want is a δ such that:

    |x - 1/2| < δ implies: |x - 1/2| < c/m. any ideas?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2011
    Posts
    12

    Re: Precise Definition of A Limit Problem

    I am lost :\. I tried to go through the usual steps of finding the precise def of a limit but it looks long and confusing. I basically got c>o < mx+b - [(m/2) +b] < c>0 but I have no idea what to do from there.
    Last edited by Mikusboi; September 16th 2012 at 11:22 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1

    Re: Precise Definition of A Limit Problem

    Quote Originally Posted by Mikusboi View Post
    I am lost :\. I tried to go through the usual steps of finding the precise def of a limit but it looks long and confusing. I basically got c>o < mx+b - [(m/2) +b] < c>0 but I have no idea what to do from there.
    Surely you see that \left| {mx + b - \left( {\frac{m}{2} + b} \right)} \right| = \left| m \right|\left| {x - \frac{1}{2}} \right|~?
    Let \delta=\frac{c}{|m|+1}>0

    If \left| {x - \frac{1}{2}} \right|<\delta then \left| {mx + b - \left( {\frac{m}{2} + b} \right)} \right|<c~.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 21st 2010, 08:46 AM
  2. Replies: 2
    Last Post: June 4th 2010, 07:02 AM
  3. I can't figure out this log problem!!!!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 2nd 2009, 04:18 PM
  4. Another Problem- Can't figure it out.
    Posted in the Statistics Forum
    Replies: 4
    Last Post: March 31st 2009, 08:19 AM
  5. Can't figure out this problem
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: January 9th 2008, 04:05 PM

Search Tags


/mathhelpforum @mathhelpforum