Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition Of Infinite Limits

  1. May 24, 2012 #1
    I have attached this definition that my book provides. My question is does that part "for each M > 0 there exists δ > 0 such that f(x) > M, mean that whenever you M close to the limit, you can find a δ that will give M1 that is closer to the limit?

    Attached Files:

    • PF.JPG
      File size:
      35.2 KB
    • PF2.JPG
      File size:
      6.2 KB
    Last edited: May 24, 2012
  2. jcsd
  3. May 24, 2012 #2
    Is there more information that is needed? Because that is all the information contained in the book.
  4. May 24, 2012 #3


    User Avatar
    Science Advisor

    No, just don't expect people to be sitting around waiting to answer your question! To get a response within 24 or 48 hours is pretty good.

    Not exactly. It means that no finite number, M, can be the limit because, for any [itex]\delta> 0[/itex] there exist x within [itex]|x- c|< \delta[/itex] such that f(x)> M.
  5. May 26, 2012 #4
    You are wrong in a couple of ways halls of ivy

    1. "no finite number is the limit." that is not correct. it is implied, but it is not the same

    2. "there is some x in the delta ball." that would just mean the function is not bounded near there. You are allowing f to jump back down as often as it wants.

    The definition needs to say that within these delta balls, all x values land above M. Choose M as large as you want, then there is such a delta ball. And all of us, even me, need to be careful about the order in which we say these things. What I said just now could easily be misinterpreted.

    So, to put it better, Pick any M value, large as you want, I dare you. Then I promise you a delta ball. You can pick any x in that delta ball, and I guarantee that x will land above M. That's why in the picture they shaded it blue, they're trying to suggest all x vaules inside will work.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook