Results 1 to 8 of 8

Math Help - Epsilon Delta Proof

  1. #1
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106

    Epsilon Delta Proof

     \lim{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}

    apparently  Delta=\frac {3Epsilon}{5}

    you see I know how they got the 5

    but for the 3 all they did was say  |x^2+x+1|>1

    and for some reason that lead them to 3

    MAT196F Midterm Test Solutions

    it's kind of hard to read but it's the first question
    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 akhayoon View Post
     \lim{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}

    apparently  Delta=\frac {3Epsilon}{5}

    you see I know how they got the 5

    but for the 3 all they did was say  |x^2+x+1|>1

    and for some reason that lead them to 3

    MAT196F Midterm Test Solutions

    it's kind of hard to read but it's the first question
    no. for the 3, they combined the fractions. that is, they subtracted 2/3 from the expression
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    I know they did that

    but they said that if  |x-1| < 1, then 0 < x < 2, so,  |2x+1| < 5

    so this give you permission to turn the |2x-1| into a five but to get rid of the

     x^2+x+1 all they said is that it's bigger than one so there

    so how does that give them permission to make the that polynomial equal 1
    ??
    Last edited by akhayoon; December 8th 2007 at 07:03 PM. Reason: typo
    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 akhayoon View Post
    I know they did that

    but they said that if  |x-1| < 1, then 0 < x < 2, so,  |2x+1| < 5

    so this give you permission to turn the |2x-1| into a five but to get rid of the

     x^2+x+1 all they said is that it's bigger than one so there

    so how does that give them permission to make the that polynomial equal 1
    ??
    i'm sorry, i don't see what you are talking about. where did they equate the polynomial to 1? they just said |x^2 + x + 1| > 1. which is okay.


    This is my 57th post!!!!
    Last edited by Jhevon; December 8th 2007 at 07:52 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these

     \frac {|x-1||2x+1|}{(3|x^2+x+1|)} < E

    |2x-1|<5 so that means that I could use 5 instead of this expression
    I know they got this from  0<x<2 sweet...

    and then they go |x^2+x+1| > 1
    but where did this come from did they use  |x-1|<1 again somehow??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    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 akhayoon View Post
    I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these

     \frac {|x-1||2x+1|}{(3|x^2+x+1|)} < E

    |2x-1|<5 so that means that I could use 5 instead of this expression
    I know they got this from  0<x<2 sweet...

    and then they go |x^2+x+1| > 1
    but where did this come from did they use  |x-1|<1 again somehow??
    yes, but not directly. they got that from 0 < x < 2. note that when x = 0, the expression |x^2 + x + 1| = 1. but x is greater than zero, so we must have |x^2 + x + 1| > 1, since we increased the expression by a positive amount

    are you still confused?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    ok thanks, that explains it, so why couldn't I just make

    |2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?

    why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    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 akhayoon View Post
    ok thanks, that explains it, so why couldn't I just make

    |2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?

    why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?
    remember, we stated out with |x - 1|< 1, which is why things played out that way. besides, if x = 0, |2x + 1| is not bigger than 1. we need to make x a range of values here, as we wanted inequalities. the actual value doesn't matter that much. we could have made |x - 1| < 1/2 or whatever. but 1 looks nice
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Delta Epsilon Proof (Given Epsilon)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 15th 2010, 03:42 PM
  2. Epsilon-delta proof
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 16th 2010, 05:21 AM
  3. delta-epsilon proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 30th 2009, 03:28 AM
  4. Replies: 0
    Last Post: October 27th 2009, 07:06 AM
  5. Delta Epsilon Proof
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 11th 2009, 03:11 PM

Search Tags


/mathhelpforum @mathhelpforum