Results 1 to 7 of 7

Thread: Epsilon-Delta

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    59

    Epsilon-Delta

    Express the following as an Epsilon-Delta proof (to show that it is continuous):

    $\displaystyle \lim_{x\rightarrow - \infty}f(x) = L$

    Can someone give me ideas for this one?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Jacobpm64 View Post
    Express the following as an Epsilon-Delta proof (to show that it is continuous):

    $\displaystyle \lim_{x\rightarrow - \infty}f(x) = L$

    Can someone give me ideas for this one?
    Continous has nothing to do with that.

    But it means.
    For any $\displaystyle \epsilon >0$
    There exists a $\displaystyle N$.
    If,
    $\displaystyle x<N$ is in the domain
    Then,
    $\displaystyle |f(x)-L|<\epsilon$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2006
    Posts
    59
    all right, I understand where you got the x < N, because x doesn't approach a specific value, but it does have to be less than some number N.

    So, usually I'd write
    $\displaystyle |f(x) - L| < \epsilon \quad \Rightarrow \quad |f(x) - L| < \epsilon$

    Then I'd try to get that in the form $\displaystyle |x - c| < \delta $

    How do I do that here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Jacobpm64 View Post
    all right, I understand where you got the x < N, because x doesn't approach a specific value, but it does have to be less than some number N.

    So, usually I'd write
    $\displaystyle |f(x) - L| < \epsilon \quad \Rightarrow \quad |f(x) - L| < \epsilon$

    Then I'd try to get that in the form $\displaystyle |x - c| < \delta $

    How do I do that here?
    There is no delta here. It is a different type of limit.
    Maybe you mean this.
    $\displaystyle \forall \epsilon >0, \exists N \, \, \, \, \, x<N\to |f(x)-L|<\epsilon$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2006
    Posts
    59
    Oh, all right.. No delta because x is extending without bound towards negative infinity, but there is an epsilon because the function approaches a finite value of L, correct?

    I understand that.

    But, I still don't know how I'd go about constructing a proof.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Jacobpm64 View Post

    But, I still don't know how I'd go about constructing a proof.
    Say you want to show,
    $\displaystyle \lim_{x\to -\infty} \frac{1}{x}=0$

    You need to show that for any $\displaystyle \epsilon>0$
    We can choose an $\displaystyle N$
    Such that if,
    $\displaystyle x<N$ is in the domain of the function.
    Implies,
    $\displaystyle \left|\frac{1}{x}\right|<\epsilon$
    Since both side are positive we can take the reciprocal of both sides, but then we have to flip the inequality, obtaining an equivalent statement,
    $\displaystyle |x|<\frac{1}{\epsilon}$
    That means, $\displaystyle x<\frac{1}{\epsilon}$.
    We also note that,
    $\displaystyle \frac{1}{\epsilon}$ is in the domain of $\displaystyle f(x)$.

    So this tell us given any $\displaystyle \epsilon$ we need to choose $\displaystyle N=\frac{1}{\epsilon}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2006
    Posts
    59
    wow, major headache.. I understand all your examples

    Thing is. The only thing that's killing me on mine is that there is no specific function... I'm just given f(x)

    In your example, you wrote $\displaystyle |f(x) - L| < \epsilon $

    as $\displaystyle x < \frac{1}{ \epsilon } $

    That isn't too bad to do because f(x) is defined as a specific function, $\displaystyle \frac{1}{x} $

    I don't know how I'd write $\displaystyle |f(x) - L| < \epsilon $ in the form x < something, when I don't have any specific function, so I only have an f(x), not an x.

    *sigh*
    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: Sep 15th 2010, 03:42 PM
  2. Replies: 0
    Last Post: Oct 27th 2009, 07:06 AM
  3. epsilon delta?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 2nd 2009, 10:09 AM
  4. Delta Epsilon
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 7th 2008, 05:33 PM
  5. epsilon delta
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Aug 11th 2008, 02:50 AM

Search Tags


/mathhelpforum @mathhelpforum