Results 1 to 9 of 9

Thread: Question concerning delta epsilon proof

  1. #1
    Junior Member
    Joined
    Dec 2010
    Posts
    31

    Question concerning epsilon-delta proof

    $\displaystyle Lim\frac{\1}{x^2}=\frac{\1}{4}$

    $\displaystyle x\rightarrow2$

    So I start with :


    $\displaystyle \frac{\1}{x^2}-\frac{\1}{4}$

    which breaks down to :

    $\displaystyle \frac{\(x-2)(x+2)}{4x^2}$

    But I'm confused how to bound this to get a delta.
    (PS: I cant figure out how to put the absolute value signs in here, but I know they are supposed to be there)

    Do I break it down to :

    $\displaystyle (x-2)*\frac{\(x+2)}{4x^2}=(x-2)*\frac{\1}{4x}+\frac{\1}{2x^2}$

    Then what?
    Last edited by Jerry99; Aug 19th 2012 at 11:31 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,743
    Thanks
    2814
    Awards
    1

    Re: Question concerning epsilon-delta proof

    Quote Originally Posted by Jerry99 View Post
    $\displaystyle Lim\frac{\1}{x^2}=\frac{\1}{4}$
    $\displaystyle x\rightarrow2$
    So I start with :
    $\displaystyle \frac{\1}{x^2}-\frac{\1}{4}$
    which breaks down to :
    $\displaystyle \frac{\(x-2)(x+2)}{4x^2}$
    But I'm confused how to bound this to get a delta.
    (PS: I cant figure out how to put the absolute value signs in here, but I know they are supposed to be there)
    [TEX]\left|{\frac{\1}{x^2}-\frac{\1}{4}\right|[/TEX] gives $\displaystyle \left|{\frac{\1}{x^2}-\frac{\1}{4}\right|$

    $\displaystyle \left|{\frac{\1}{x^2}-\frac{\1}{4}\right|=\frac{|x-2||x+2|}{|4x^2|}$

    If $\displaystyle 0<\delta<1~\&~0<|x-2|<\delta$ then $\displaystyle |x+2|<5$ and $\displaystyle \left|{\frac{\1}{4x^2}\right|<\frac{1}{4}$

    See what you can do with that.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2010
    Posts
    31

    Re: Question concerning epsilon-delta proof

    I get how $\displaystyle \left|x+2\right|<5$

    But how do you get the $\displaystyle \left|\frac{\1}{4x^2}\right|<\frac{\1}{4}$

    and thanks for the absolute value sign tip
    Last edited by Jerry99; Aug 19th 2012 at 12:03 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2010
    Posts
    31

    Re: Question concerning delta epsilon proof

    I think I get what is going on now actually. that has to be less than 1/4 basically

    thanks for the help
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,743
    Thanks
    2814
    Awards
    1

    Re: Question concerning epsilon-delta proof

    Quote Originally Posted by Jerry99 View Post
    I get how $\displaystyle \left|x+2\right|<5$
    But how do you get the $\displaystyle \left|\frac{\1}{4x^2}\right|<\frac{\1}{4}$
    If $\displaystyle 1<x<3$ then $\displaystyle 4<4x^2<36$ or $\displaystyle \frac{1}{36}<\frac{1}{4x^2}<\frac{1}{4}$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2010
    Posts
    31

    Re: Question concerning epsilon-delta proof

    Ok cool I get it. Thanks for the help
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2010
    Posts
    31

    Re: Question concerning delta epsilon proof

    Sigh , well, I thought I did. What am I missing here that allows you to make the statement of $\displaystyle 4 < 4x^2<36 =\frac{\1}{36}<\frac{\1}{4x^2}<\frac{\1}{4} $


    My min $\displaystyle \delta$from this : 1, $\displaystyle \frac{\epsilon}{3},\frac{\epsilon}{5}$ ??
    Last edited by Jerry99; Aug 19th 2012 at 02:01 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,743
    Thanks
    2814
    Awards
    1

    Re: Question concerning delta epsilon proof

    Quote Originally Posted by Jerry99 View Post
    Sigh , well, I thought I did. What am I missing here that allows you to make the statement of $\displaystyle 4 < 4x^2<36 =\frac{\1}{36}<\frac{\1}{4x^2}<\frac{\1}{4} $
    My min $\displaystyle \delta$from this : 1, $\displaystyle \frac{\epsilon}{4},\frac{\epsilon}{5}$ ??
    $\displaystyle \begin{array}{l}1 < x < 3\\ 1 < {x^2} < 9\\ 4 < 4{x^2} < 36\\ \dfrac{1}{{36}} < \dfrac{1}{{4{x^2}}} < \dfrac{1}{4} \end{array}$
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    Re: Question concerning delta epsilon proof

    if 0 < a < b < c, then 0 < a2 < ab < b2 < bc < c2, or in short:

    0 < a2 < b2 < c2.

    similarly, if 0 < a < b < c then:

    0 < 1/c < 1/b < 1/a

    the basis for the last inequality is this:

    we define for two fractions a/b, c/d: a/b < c/d if and only if ad < bc. note this agrees with our usual notion if b = d = 1.

    this means that if a < c, that when we compare:

    1/a and 1/c, we have 1(c) > a(1), so 1/a > 1/c.

    for example: 4 is bigger than 2, so 1/4 is smaller than 1/2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Epsilon - Delta Proof Question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Oct 10th 2011, 06:31 PM
  2. Question regarding above epsilon-delta proof
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Oct 27th 2010, 11:38 AM
  3. Solving Delta Epsilon Proof (Given Epsilon)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 15th 2010, 03:42 PM
  4. Replies: 0
    Last Post: Oct 27th 2009, 07:06 AM
  5. Delta-Epsilon Proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Aug 6th 2009, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum