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Precise definition of the limit (epsilon-delta)

  1. Jun 26, 2015 #1
    Sorry, I am really struggling with the precise definition of the limit. I have a specific question I'm trying to work out
    lim(x->2) (4x2+2)=18

    skipping the introduction part
    any advice? I am just not sure how to get rid of the 2 value to re-arrange |(4x2+2)-18| to look like |x-2|

    |x-2|<delta |(4x2+2)-18|<epsilon
    |(4x2-16|<epsilon

    I did go down the route of
    |(4x2+2)-18| = |2x+4| |2x-4|<epsilon
    but this step just lead me to the same problem
     
  2. jcsd
  3. Jun 26, 2015 #2

    HallsofIvy

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    So you realize that you have [itex]|2x- 4||2x+ 4|< \epsilon[/itex]. You should be immediately able to see that this is the same as [itex]4|x- 2||x+ 2|< \epsilon[/itex] and then [itex]|x-2||x+2|< \epsilon/4[/itex].

    And you want "[itex]|x- 2|< \delta[/itex]". Now, if that "|x+ 2|" were a positive constant, you could just divide by it: [itex]|x- 2|< \epsilon/(4|x+2|)[/itex] but because it contains "x" you can't do that. What you can do is replace |x+2| with an constant upper bound. If we could say that |x+ 2|< U, for x close to 2, then 1/U< 1/|x+2| and [itex]\epsilon/(4U)< \epsilon/4|x+ 2|)[/itex].

    If x is close to 2, we can certainly say that -1< x- 2< 1 (choosing -1 and 1 just because they are easy) so that 4- 1= 3< x+ 2> 4+ 1= 5. An upper bound on x+ 2 is 5.
     
  4. Jun 26, 2015 #3
    Not sure if this is right but thought I'd post my progress along with helpful sites in case somebody else has the same problem
    http://math.stackexchange.com/quest...-where-i-let-delta-be-a-minimum-of-two-values
    http://mathforum.org/library/drmath/view/53738.html
    (not sure on the forums policy on links, so apologies if this isnt right)

    lim(x->2) 4x^2+2) = 18

    0<|x-2|<delta

    |(4x^2+2)-18|<epsilon
    |(4x^2-16|<epsilon
    |2x+4|.|2x-4|<epsilon
    /2
    |x+2|.|x-2|<epsilon/2
    1<|x+2|<3
    -4
    -3<|x-2|<-1
    |x-2|<-1

    |x+2|.|x-2|<delta.-1
    hence
    epsilon = (-)delta ~ (-)epsilon = delta
     
  5. Jun 26, 2015 #4

    RUber

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    Your claim is that given an epsilon>0, then there is a delta>0 such that if |x-2|< delta, |(4x^2+2) - 18| < epsilon.
    You have |x-2| < delta and above you wrote |x+2| |x-2| < epsilon/2.
    HInt: ##|x-2 + 4 | \leq | x-2| + 4##.
    Solve for delta in terms of epsilon.
    Also...I disagree with where you wrote 1< |x+2|< 3 since the assumption is that x is getting close to 2.
     
  6. Jun 26, 2015 #5
    That's the part I don't truely understand to be honest. The guide I read suggested as your trying to stay witching a small range 1 either side of the value (2) |x+2| is this wrong or have I (the more likely) misunderstood
    Thanks but I don't quite get your hint
     
  7. Jun 26, 2015 #6

    Fredrik

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    They're probably trying to use the fact that if ##\delta<1## and ##|x-2|<\delta##, then |x+2| is less than...what? You can probably figure that out for yourself. This simplifies the problem. The idea here is that we don't need to find the largest ##\delta## that works, so we might as well use that freedom to choose a ##\delta## that's small enough to simplify the calculations.

    RUber is using the rewrite x+2=x-2+4 and the triangle inequality ##|x+y|\leq|x|+|y|## to find an inequality of the form ##|x+2|\leq f(\delta)## (where f is some function) that holds when ##|x-2|<\delta##. (It's up to you to figure out what f is). Then we can say that for all x such that ##|x-2|<\delta##, we have
    $$|4x^2+2-18|=4|x+2||x-2|<\text{what?}$$ Since you need this to be less than ##\varepsilon##, you can find the largest ##\delta## that works by solving the equation ##\text{what?}=\varepsilon## for ##\delta##.
     
  8. Jun 26, 2015 #7
    Why is this 4? Is it not 2 or am i seeing this wrong?

    less than 1?

    I'm struggling with the, almost, canceling down from factoring to
    [itex]|x-2||x+2|< \epsilon/2[/itex]
    from this step I get really lost, as far as I'm aware the |x-2| is controlled by delta so i have to work with |x+2|
    by this i use the fact i need to stay close to 2 so taking 1 either side i can say 1<|x+2|<3 (either side of the 2) and to make it into |x-2| i minus 4 giving -3<|x-2|<-1
    which gives me -1 which is the upper? so as
    |x-2| |x+2|<epsilon/2
    |x-2| |x+2|<delta . epsilon/-2

    now I just feel lost. I'm really not sure where I am going wrong
    thanks for the help though
     
  9. Jun 26, 2015 #8

    PeroK

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    First, look at what you've done. That's a very basic error in using the absolute value.

    Perhaps this epsilon-delta thing looks like some sort of mathematical black magic; whereas, it is simply algebra and estimation. To try to get a grip on what is happening here, I suggest you calculate the function values for ##|x-2| < \delta## for some specific values of ##\delta##.

    Note first that ##f(x) = 4x^2 + 2## is increasing for ##x > 0##. So, first we assume that ##\delta < 2## so that ##x## is always positive.

    For ##\delta = 1##, we have ##|x-2| < 1 \ \Rightarrow \ 1 < x < 3 \ \Rightarrow \ f(1) < f(x) < f(3) \ \Rightarrow \ 6 < f(x) < 38 \ \Rightarrow \ |f(x) - 18| < 20##

    Now try with ##\delta = 1/2## and ##\delta = 1/4##.

    At least this might give you an idea of what's going on. The next step, of course, is to try to work out how small ##\delta## has to be to keep ##|f(x) - 18| < \epsilon##, where ##\epsilon## is an arbitrary number.

    It strikes me, looking at your posts, that you haven't really understood how the algebraic steps with general epsilon and delta relate to the simple functional evaluations for various values of x.

    In fact, you could probably gain some insight here by drawing a graph of the function as well, so you can see what's going on when you keep x "close" to 2.
     
  10. Jun 26, 2015 #9

    HallsofIvy

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    I am not sure what you are seeing but you have |2x+ 4||2x- 4|= |2(x+ 2)||2(x- 2|= 2|x+ 2|(2)|x- 2|= 4|x+ 2||x- 2|.

    You just said you "have to work with |x+ 2|" so why are you looking at |x- 2|? We want to be able to say that "if [itex]|x- 2|< \delta[/itex]- that is if x is close enough to 2 then [itex]|(x- 2)(x+ 2)|< \epsilon[/itex]. In particular we can choose to look at x between 1 and 3- so that x is at least that "close to 2". (x, NOT x+ 2, between 1 and 3) then since 1< x< 3, 3< x+ 2< 5.

    You appear to be doing things backwards! Surely you see that if 1< x+ 2< 3 then -1< x< 1 so x is not anywhere near the "target" of x going to 2?

     
  11. Jun 26, 2015 #10

    Fredrik

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    You said that something about "...trying to stay witching a small range 1 either side of the value (2)". That sentence isn't easy to understand, but I interpreted it as keeping x at a distance less than 1 from the number 2, i.e. as requiring that |x-2|<1.

    The number 1 is just a convenient choice that makes it easy to handle the |x+2| factor.

    The thing that you need to be less than ##\varepsilon## is
    $$|4x^2+2-18|=|4x^2-16|=4|x^2-4|=4|x+2||x-2|,$$ so I think you're off by a factor of 2.

    The fact that |x-2| is "controlled by delta" ensures that |x+2| is too. You can see this by rewriting |x+2| as |(x-2)+4| and using the triangle inequality. A more intuitive (but not as rigorous) way is to draw a line to represent the real numbers, and mark the numbers -2 and 2. |x-2| is the distance between x and 2. |x+2| is the distance between x and -2. If x is at a distance from 2 that's less than ##\delta##, then what is the least upper bound on the distance between x and -2?
     
  12. Jun 26, 2015 #11

    verty

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    This looks like a textbook example of what this part of the forum is not supposed to be used for, because all I see are questions on questions, all of the type "what do you mean by this, what do you mean by that?" Cal124's last post is exactly like this. That thing about 4 versus 2, I mean come on.

    Cal124 must put the effort in. These limit problems are not hard but I don't see any serious thought being spent on it by him and for that reason, I think it is in the wrong place and the help has been plentiful but taken for granted, so far as I can tell.
     
  13. Jun 26, 2015 #12
    Sounds like maybe this forum isn't for me then, maybe when I'm better educated. Nothing was taken for granted, thanks to those who did help, I will study your advice and hopefully get it.
     
  14. Jun 26, 2015 #13

    Fredrik

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    We don't need you to be better educated. We just need you to do a bigger part of the work towards a solution. We're supposed to give you hints so that you can solve the problem for yourself, not solve the problem for you. But what happened here was that we gave you a piece of the solution, and then you asked questions until you got the next piece, and so on. This isn't just your mistake. Since you're new here, I'd say that we are more to blame for this than you.
     
  15. Jun 27, 2015 #14

    verty

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    I apologize for jumping in so brashly. You can put it down to me having a bad day.
     
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