Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Gusbob

Math Help - Epsilon delta proof for a limit.

  1. #1
    Junior Member Bradyns's Avatar
    Joined
    Apr 2013
    From
    Maitland, NSW, Australia
    Posts
    39
    Thanks
    12

    Epsilon delta proof for a limit.

    Just had an exam where the question was roughly:

    =================================================
    Using an epsilon delta definition of limit, prove that the limit

    \lim_{x\rightarrow 1}x^2 + 3 = 4
    =================================================

    I am not very confident with proving the limit definition of quadratics..
    So I folded and decided to do it by two sided limit.

    Rough attempt
    For an \epsilon > 0, there exists a \delta > 0, such that |x-a|<\delta\Rightarrow |f(x)-L|<\epsilon:

    Left hand limit:
    1 - \delta < 1 \Rightarrow \delta > 0

    4 - \epsilon < 4 \Rightarrow \epsilon > 0

    Right hand limit:
    1 < 1 + \delta \Rightarrow \delta > 0

    4 < 4 + \epsilon \Rightarrow \epsilon > 0

    There exists an \epsilon > 0 and a \delta > 0 on both sides of 1, which approach the value of 4.

    Thefore the limit does exist, and it is 4.

    Pretty sure I messed up monumentally.. Linear functions are so much easier to prove than quadratics. :/
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87

    Re: Epsilon delta proof for a limit.

    Erm...I'm not sure what you're doing here. It seems to be a bit of nonsense sorry - you don't get to choose \epsilon. I'll show you how to do it properly.

    Claim: For a fixed \epsilon >0, there exists a \delta>0 such that |x-1|\leq \delta \implies |x^2+3-4|\leq \epsilon

    Such a \delta, if it exists, must satisfy the following inequality:

    Now |x^2-1|=|(x-1)(x+1)|\leq |x-1|\cdot |x+1|\leq \delta |x+1|

    Obviously we can't actually bound |x+1| for arbitrary x, but we don't actually care about what happens to x far away from 1. For example, we can require \delta \leq 1. Then |x-1|<\delta \leq 1 \implies 0\leq x\leq 2 \implies 1\leq |x+1| \leq 3. For such \delta, we have

    |x^2-1| \leq \delta |x+1| \leq 3\delta. If we want the last expression to be \epsilon, we require \delta  \leq \frac{\epsilon}{3} .



    To recap, we have
    |x^2-1|\leq 3\delta \leq \epsilon

    with \delta subject to the conditions \delta\leq 1 and \delta \leq \frac{\epsilon}{3}. Choosing \delta = \min\{1,\frac{\epsilon}{3}\} satisfies these conditions. As such you have found your desired \delta.



    Writing the proof out formally:

    Fix \epsilon >0 and choose \delta =\min\{1,\frac{\epsilon}{3}\} . Suppose |x-1|\leq \delta. Then

    |x^2+3-4|
    =|x^2-1|
    \leq |x-1|\cdot |x+1|
    \leq \frac{\epsilon}{3} |x+1| since  |x-1| < \delta \leq \frac{\epsilon}{3}
    \leq \frac{\epsilon}{3}\cdot 3 since  |x-1| < \delta \leq 1 \implies |x+1|\leq 3
    \leq \epsilon
    Thanks from Bradyns
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: Epsilon delta proof for a limit.

    There's a good explanation of how to write these epsilon-delta proofs - it's one of the four "sticky" threads on the Calculus forum here at MHF.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Epsilon delta limit proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 16th 2012, 12:24 AM
  2. epsilon delta proof of limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 12th 2012, 07:09 AM
  3. epsilon delta limit proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 9th 2010, 10:24 AM
  4. Epsilon Delta Proof of a Limit
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 21st 2009, 08:37 PM
  5. Epsilon-Delta Limit Proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 13th 2006, 06:52 AM

Search Tags


/mathhelpforum @mathhelpforum