Results 1 to 3 of 3

Math Help - ε-δ Limit with an x term?

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    43

    ε-δ Limit with an x term?

    Hello , I have been having trouble with understanding what to do with those ε-δ Limits that involve an extra "x" factor in the result, I'll just write out the problem and perhaps you could correct or instruct me on the logic behind it.

    \lim_{x \to 2}(x^2  +  x  - 2)  =  4

    0 < |x - 2| < \delta \Rightarrow |(x^2 + x - 2) - 4| < \epsilon

    0 < |x - 2| < \delta \Rightarrow |x^2 + x - 6| < \epsilon

    0 < |x - 2| < \delta \Rightarrow |(x + 3)(x - 2)| < \epsilon

    0 < |x - 2| < \delta \Rightarrow |x + 3||x - 2| < \epsilon

    Here, I'm kind of confused. I've seen the best explanation so far here: Created with Camtasia Studio 5 & I'd like to follow the method to get everything in working order.

    1: Because of the |x - 2|< δ, (i.e. the x-->2 in the original limit), we know that x is very close to 2, just a bit bigger than it. Because this is so the |x + 3| term is very close to 5, just a bit above it. In fact, it will definitely be less than 6.

    We can say with confidence that;

    |x + 3||x - 2| < 6|x - 2|

    2: Here I'm a bit unsure, can we say that;

    6|x - 2| < ε

    |x - 2|< ε/6 ? I think this works because although the gap from. say, 5.0000001 to being less than 6 (from the inequality) is big, when we divide ε by 6 we are creating more space.

    3: Is there a way to make it look pretty? Is the answer that as long as δ = ε/6 we can guarantee the limit?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    I recommend you read this for detailed explanation and working.

    http://www.mathhelpforum.com/math-he...ta-proofs.html

    You'll find example 3 to be especially helpful.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    43
    I already had that pdf on my laptop & I never understood it until I followed the video I linked to but I may understand it now.

    If I take my example;

    |x + 3||x - 2| < \epsilon

    |x - 2| < \frac{ \epsilon}{ |x + 3|}

    What I can do is look back at |x - 2| & if I want to bound it within 1 unit it would be equivalent to bound the |x + 3| factor in the denoinator within a distance of 1 as well;

    |x - 2| < 1

    -1 < x - 2 < 1

    Then, because the limit is going near 2 we'll take the |x + 3| and realise it's near 5, but because of the constriction in |x - 2| of being within 1 unit we'll do the same for the |x + 3|

    -1 < x - 2 < 1 \Rightarrow  4 < x + 3 < 6

    Then, if we take 6 as the min. i.e. δ = (1,ε/6)

    |x + 3||x - 2| < \epsilon

    6|x - 2| < \epsilon

    |x - 2| < \frac{ \epsilon}{6}

    |x - 2| < \frac{ \epsilon}{6}

    So;

    |x + 3||x - 2|

    6|x - 2|

    6 ( \frac{ \epsilon}{6})

    \epsilon


    I haven't made any careless mistakes or assumptions have I?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Integration term by term and orthogonality
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 23rd 2011, 04:36 PM
  2. Replies: 1
    Last Post: February 19th 2011, 12:21 PM
  3. Power series, term-by-term integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 8th 2010, 03:14 AM
  4. [SOLVED] Ratio of the second term to the first term
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: September 26th 2008, 07:13 AM
  5. Replies: 7
    Last Post: August 31st 2007, 09:18 PM

Search Tags


/mathhelpforum @mathhelpforum