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Math Help - Give the epsilon delta definition

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    Senior Member vaironxxrd's Avatar
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    Give the epsilon delta definition

    Hello Everyone,

    I'm reading a section of my book and ran into a couple of problems I can't wrap over my head.
    My book does explain the epsilon-delta concept, but I can't understand how to apply it.

    I have the following problem: Give the  \epsilon - \delta  definition for \lim_{u \to a} g(u) = m

    My attempt to solve this problem.

    0 < |x-c| < \delta
    |g(u)-L | < \epsilon

    This is not so much of an attempt, but rather more of what I know about it. I'm still browsing websites to understand how this works.
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    Re: Give the epsilon delta definition

    Quote Originally Posted by vaironxxrd View Post
    Hello Everyone,

    I'm reading a section of my book and ran into a couple of problems I can't wrap over my head.
    My book does explain the epsilon-delta concept, but I can't understand how to apply it.

    I have the following problem: Give the  \epsilon - \delta  definition for \lim_{u \to a} g(u) = m

    My attempt to solve this problem.

    0 < |x-c| < \delta
    |g(u)-L | < \epsilon

    This is not so much of an attempt, but rather more of what I know about it. I'm still browsing websites to understand how this works.
    I think you need more "words" to clarify what those symbols mean. That is instead of
    0< |x- c|< \delta
    |g(u)- L|< \epsilon

    You should say: "For any number \epsilon> 0 there exist a number \delta> 0 so that if |x- c|< \delta,
    then |g(u)- L|< \epsilon."

    That is, we can make g(u) as close L as we like (within distance \epsilon) by making x close to c (within distance \delta).
    Thanks from topsquark and vaironxxrd
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    Re: Give the epsilon delta definition

    If you're not really understanding the \displaystyle \begin{align*} \delta - \epsilon \end{align*} definitions of a limit, it might help with a metaphor.

    When I do \displaystyle \begin{align*} \epsilon - \delta \end{align*} proofs, I think of myself pulling pizzas out of an oven (I used to work in a pizza shop). Think of there being an "ideal" level of cooking for your pizza. Obviously, it is not going to be possible to get this "ideal" amount of cooking for every pizza (or possibly even any pizza), but there is a certain "tolerance" you can have for over-cooking or under-cooking before you consider it raw or burnt. As long as you are reasonably close to the right amount of time needed, then your level of cooking will be considered acceptable. Then as you gain more experience, you should be able to get closer and closer to keeping the pizzas in the oven for the ideal amount of time, thereby making your pizzas closer and closer to the ideal level of cooking, which means you would expect that your tolerance would decrease as you'd be getting used to your pizzas being cooked properly.

    So if we were to call the amount of time in the oven \displaystyle \begin{align*} x \end{align*}, then the level of cooking is some function of x \displaystyle \begin{align*} f(x) \end{align*}. We said there is an ideal level of cooking, we could call that \displaystyle \begin{align*} L  \end{align*}, which means there is a point in time \displaystyle \begin{align*} x = c \end{align*} which gives this ideal level of cooking. Remember we said that as long as we have kept the pizzas in the oven for an amount of time reasonably close to \displaystyle \begin{align*} c \end{align*}, say \displaystyle \begin{align*} \delta \end{align*} units of time away from it, then our level of cooking would be considered acceptable, or within some tolerance which we could call \displaystyle \begin{align*} \epsilon \end{align*}. So we need to show that \displaystyle \begin{align*} \delta \end{align*} and \displaystyle \begin{align*} \epsilon \end{align*} are related, so that you are guaranteed that as you get experience and keep your pizzas in the oven closer to the right amount of time ( i.e. \displaystyle \begin{align*} \delta \end{align*} gets smaller) then so will your tolerance \displaystyle \begin{align*} \epsilon \end{align*} get smaller and closer to the ideal level of cooking.

    Do you see now what it means to show \displaystyle \begin{align*} 0 < |x - c| < \delta \implies |f(x) - L | < \epsilon \end{align*}? It means if you have set a tolerance around your ideal limiting value, then as long as you are reasonably close to \displaystyle \begin{align*} x = c \end{align*}, then you are guaranteed that your function value is within your tolerance away from the limiting value, and by showing the relationship between \displaystyle \begin{align*} \delta \end{align*} and \displaystyle \begin{align*} \epsilon \end{align*}, you are guaranteed that as your \displaystyle \begin{align*} \delta \end{align*} gets smaller and you close in on \displaystyle \begin{align*} x = c \end{align*}, then your tolerance will get smaller and your \displaystyle \begin{align*} f(x) \end{align*} will close in on \displaystyle \begin{align*} L \end{align*}.
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    Re: Give the epsilon delta definition

    Quote Originally Posted by topgear2015 View Post
    In this link you will find every math solution step by step with instructions!!!
    Step by step Calculus Math Problem Solutions - Your math Solver
    I think the mods should delete this post. It is advertising material and also goes against the policies of the site which is that we believe students should be doing their own work!
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    Re: Give the epsilon delta definition

    Quote Originally Posted by Prove It View Post
    I think the mods should delete this post. It is advertising material and also goes against the policies of the site which is that we believe students should be doing their own work!
    In case the post is deleted I will save the first two answers given. As you might know I always come here for guidance and not for the answer. I think your metaphor gave a much clear definition from that in my book. Now I'll have to solve this problem and others to see if I truly understand the concept.
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    Re: Give the epsilon delta definition

    Quote Originally Posted by HallsofIvy View Post
    I think you need more "words" to clarify what those symbols mean. That is instead of.....
    Quote Originally Posted by Prove It View Post
    If you're not really understanding the \displaystyle \begin{align*} \delta - \epsilon \end{align*} definitions of a limit, it might help with a metaphor.....
    I believe I have more knowledge about this concept, I will try to solve some problems and hopefully I actually got the message.

    Prove using an \epsilon - \delta argument that \lim_{x \to 3} 2x+1 = 7

    My solution:

    a = 3, L = 7, f(x) = 2x+1

    |f(x) - L | < \epsilon=

    |(2x+1)-7|<\epsilon=

    |2x-6|<\epsilon =

    2|x-3|<\epsilon =

    |x-3|<\epsilon/2=

    \delta = \epsilon/2

    After reading this document I feel like I still haven't completed the proof.

    http://www.ocf.berkeley.edu/~yosenl/...ilon-delta.pdf
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    Re: Give the epsilon delta definition

    Quote Originally Posted by vaironxxrd View Post
    Prove using an \epsilon - \delta argument that \lim_{x \to 3} 2x+1 = 7

    My solution:
    a = 3, L = 7, f(x) = 2x+1

    |f(x) - L | < \epsilon=

    |(2x+1)-7|<\epsilon=

    |2x-6|<\epsilon =
    2|x-3|<\epsilon =

    |x-3|<\epsilon/2=

    \delta = \epsilon/2
    You did a perfectly reasonable proof. Why doubt yourself?
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    Re: Give the epsilon delta definition

    Quote Originally Posted by Plato View Post
    You did a perfectly reasonable proof. Why doubt yourself?
    If you see the Berkeley website on the first example they expanded more afterwards.
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    Re: Give the epsilon delta definition

    Quote Originally Posted by vaironxxrd View Post
    If you see the Berkeley website on the first example they expanded more afterwards.
    But that is purely a matter of style. What you did is fine.
    If you were in a particular class (or with a particular instructor) then yes, you should follow the style demands of that situation.
    But as I understand what you are doing, that is not your case.
    Last edited by Plato; September 18th 2013 at 04:08 PM.
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    Re: Give the epsilon delta definition

    Quote Originally Posted by vaironxxrd View Post
    I believe I have more knowledge about this concept, I will try to solve some problems and hopefully I actually got the message.

    Prove using an \epsilon - \delta argument that \lim_{x \to 3} 2x+1 = 7

    My solution:

    a = 3, L = 7, f(x) = 2x+1

    |f(x) - L | < \epsilon=

    |(2x+1)-7|<\epsilon=

    |2x-6|<\epsilon =

    2|x-3|<\epsilon =

    |x-3|<\epsilon/2=

    \delta = \epsilon/2

    After reading this document I feel like I still haven't completed the proof.

    http://www.ocf.berkeley.edu/~yosenl/...ilon-delta.pdf
    No you haven't completed the proof. Now that you've shown the existence of such a \displaystyle \begin{align*} \delta \end{align*}, you then need to show the implication \displaystyle \begin{align*} 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon \end{align*}. Thankfully, this usually is just writing down your scratch-work steps in reverse
    Thanks from vaironxxrd
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