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A function 'continous' at a 'point'.

  1. Oct 4, 2009 #1
    I was wondering that the continuity of a function is an actually function of these δs...if they are large they might cover values where the function is broken...so things actually depend on these δs and not c if δ is not infinitely small.
  2. jcsd
  3. Oct 4, 2009 #2


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    ?? :confused:

    that's not the way it works …

    ε is given, not δ …

    for a given ε, you're allowed to choose a δ as small as you like. :smile:
  4. Oct 4, 2009 #3
    Man...you have to agree this statement is very badly written...I can't understand a thing (now

    Can someone explain??
  5. Oct 4, 2009 #4
    Yes also, suppose in the continuous function f(x) can it happen that for an infinity small change in x the value of f(x) does not change?
  6. Oct 4, 2009 #5
    if you want to examine a limit as x->a, f(a) is the limit, so you want to pick an interval of points around that limit.. say, an interval of 1. and then there will exist corresponding x values for that interval. if you picked the delta first, who knows what you'll get?

    if you ever think about how "for every epsilon there exists a delta" works, think about a function every y there exists x such that ƒ(x) = y
  7. Oct 4, 2009 #6


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    The idea behind saying that

    |f(c+h) - f(c)| < \epsilon \qquad \text{ for } |h| \le \delta

    is that when you are "really close" to c, function values are "really close" to f(c).

    In general (not always) the smaller the value of [tex] \epsilon [/tex] you select, the smaller must be [tex] \delta [/tex]

    Think about this geometric approach. Draw a portion of an arbitrary continuous function (draw any continuous curve) - for a specific example, draw it near [tex] c =2 [/tex], and suppose [tex] f(2) = 5 [/tex].

    Now pick [tex] \epsilon = 0.05 [/tex] and draw the two horizontal lines [tex] y = 5 - \epsilon[/tex] and [tex] y = 5 + \epsilon [/tex].
    Now draw two vertical lines, one on each side of [tex] c = 2[/tex] (equally spaced) so that
    all of the graph between these two lines is between the two horizontal lines you drew at the first step. The common distance these vertical lines are from 2 is [tex] \delta [/tex] - so, given [tex] \epsilon = .05 [/tex], you've just found a [tex] \delta > 0 [/tex] such that

    |f(2+h) - f(2)| < \epsilon \qquad \text{ for } |h| < \delta

    Note: I've just made an example with Latex and a few lines of tikz code. It's attached.

    Attached Files:

  8. Oct 5, 2009 #7
    I absolutely do not know limits...but I did understand what you said (sorta).

    I find the variable δ very reluctant...why not skimpily manipulate the value of h?...why make the relation h = δ and THEN manipulate h?

    So over all this |f(c+h) - f(c)| < ε && |h|<δ mechanism is trying to say that for an infinitely small change in value of h in the function f(x+h); the value of f(x+h) will also change by an infinity small value within the interval [a, b] (corresponding to value x+h) if the function is called continuous within [a, b]...it's a test for continuity.

    By this we can also define the continuity at the 'point' f(x+h) cause here the value of h is infinitely small...so WHY h, δ, x, ε........SO many variables?

    Anyway...I do not even understand the definition of the variables ε...it's neither f(x) nor f(x-h)...what is it (theoretically)?
  9. Oct 5, 2009 #8


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    Using δ instead of h has three purposes
    1)Greek letters are really cool and fun to write.
    2)δ>|h| so δ is a ristriction on h not its value
    3)δ establishes a short hand that, one knows what it means in different situations.

    Infinitely small is a strange term, we have a better one infinitesimal.
    h, δ, ε are note infinitesimal they are small enough.
    In words the definition of a limit L=lim_x->a f(x) would be
    "The function f can be made as close to L as desired by chosing x sufficiently close to a"
    or for f is continuous at a
    "The function f can be made as close to f(a) as desired by chosing x sufficiently close to a"

    ε the maximum allowed difference between the function and the limit

    Think of it like a traffic law. Your driving around in you facy function f and you see a sign "|f(x+h)-f(x)|<ε". If f is contimuous you can manage this by making sure |h|<δ so you will not get a ticket.
  10. Oct 5, 2009 #9
    Latex is not working on your signature.

    I'm inept of spelling 'infinitesimal'...my tongue sorta grows a knot in it if I try to speak it; so I prefer listening/reading it rather than spelling/typing it.

    Man...I'm not getting a thing...problem is I do not know the definitions...like what is h; for starters...I know it's not an arbitrary which, to me, seems very likely...so can someone please tell.
  11. Oct 5, 2009 #10
    So my guess was right...that complex 4 letter definition meant the same thing...thanks to recoll, I got this ebook.
  12. Oct 5, 2009 #11
    But finally I realized that the ε and δ are pretty much standard...so I should know about them.

    So what is h?
  13. Oct 5, 2009 #12


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    h is just a number smaller than delta.

    What the definition of continuity is saying is:

    If I give you some small value epsilon, you can give me a small value of delta so that if y is within delta of x, f(y) is within epsilon of f(x). It's like a game... your goal is to show something is continuous by always giving me a delta back when I give you epsilon
  14. Oct 6, 2009 #13
    And what is delta?
  15. Oct 6, 2009 #14


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    ε is a circle (or "neighbourhood") in the image space, δ is a circle (or "neighbourhood") in the object space, and h is anything inside δ.

    So, for any circle in the image space, you have to find a circle in the object space whose image lies completely inside it. :wink:
  16. Oct 6, 2009 #15
    Image and object space???? :cry:..I don't get a thing.

    Ok, this is what I pondered out -

    Most probably this is wrong cause I found the need of another variable, δ unnecessary.
    Last edited: Oct 6, 2009
  17. Oct 6, 2009 #16


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    h is not computed with \epsilon in mind. h is arbitrary, what we are specifying is that there is a continuous range from -h to h inclusive by which the distance of f(x+h) from f(x) is always less than \epsilon. We are interested in the behavior of the function locally about x, we do not care what h is, we only care about the restrictions on h and the restrictions on the change in the function when we deviate by some h.
  18. Oct 6, 2009 #17
    Thanks A LOT man...that helped my by A LOT...finally something clear that I can understand.

    So we can say that |f(x+h) - f(x)| and |f(x-h) - f(x)| should not exceed ε and ε is the constant here. Since |f(x+h) - f(x)| and |f(x-h) - f(x)| is also a function of h, ε poses a restriction on h.

    But this is sorta defining the limit...not continuity; what do we say for continuity?
  19. Oct 6, 2009 #18


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    An equivalent definition of continuity at a point is f(x) is continuous at c if[tex] \lim_{x \rightarrow c} f(x) = f(c)[/tex]

    Can you see how this matches up with the epilon delta definition?
  20. Oct 6, 2009 #19
    No, cause I do not understand the notion of limit.

    Can you please tell me in terms of ε and δ?
  21. Oct 7, 2009 #20
  22. Oct 8, 2009 #21


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    You keep saying you "do not know limits". How can you hope to understand, or even ask about, continuity, then?

    But since you ask: "[itex]\lim_{x\to a} f(x)= L[/itex]" if and only if, given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex], then [itex]|f(x)- L|< \epsilon[/itex].

    That basically means, as just about every other response here has said, that you can f(x) as close to L as you like just by taking x close enough to a.

    In order that a function, f(x), be continuous at x= a, three things must be true:
    1) f(a) must exist.
    2) [itex]\lim_{x\to a} f(x)[/itex] must exist.
    3) [itex]\lim_{x\to a} f(x)[/itex] must be equal to f(a).

    Since just writing [itex]\lim_{x\to a} f(x)= f(a)[/itex] pretty much implies that the two sides exist, most of the time we just write that.
  23. Oct 8, 2009 #22


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    You stated that [tex] f [/tex] is continuous at [tex] c [/tex] if, for every [tex] \epsilon > 0 [/tex], there is a [tex] \delta > 0 [/tex] such that

    | f(c+h) - f(c)| < \epsilon \quad \text{ if } |h| < \delta

    An intuitive statement of continuity at a point is this:

    The function [tex] f [/tex] is continuous at a number [tex] c [/tex] if [tex] f(c+h)[/tex] is
    close in value to [tex] f(c) [/tex] when [tex] c+h [/tex] is close to [tex] c [/tex].

    • [tex] \epsilon [/tex] shows how close you want [tex] f(c+h) [/tex] to be to [tex] f(c) [/tex]
    • [tex] \delta [/tex] measures how close to [tex] c [/tex] you need to pick other x-values to make the previous point come true
    • [tex] h [/tex] measures the distance the new x-value is from the number [tex] c [/tex]
  24. Oct 12, 2009 #23
    That's cause in every book first they explain continuity, then limit.

    PERFECT answer man...just perfect...thanks.
  25. Oct 12, 2009 #24


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    In every book? Peculiar- I have never seen such a book!

    PERFECT answer man...just perfect...thanks.[/QUOTE]
    Yes, it was.
  26. Oct 13, 2009 #25
    Yes, check out these books -

    Introduction to calculus - KAZIMIERZ KURATOWSKI
    Calculus - Benjamin Crowell

    Actually there are no books where limits is discussed before continuity.

    So the value of h has 2 restrictions...one ε and the other δ.

    Is the value of δ arbitrator?
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