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Prove lim(x->3) x^2 = 9 by defintion

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    [TEX][|x^2-9|< \epsilon [/TEX] if [tex]0 < |x-3| < \delta[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Hello!

    I know how to solve this task but I have certain things that I do not understand.

    Solution.

    [TEX][|x^2-9|< \epsilon [/TEX] if [tex]0 < |x-3| < \delta[/tex]

    Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

    [tex]|x+3||x-3|< \epsilon[/tex] if [tex]0<|x-3|<\delta[/tex]

    Using the triangle inequality:

    [tex]|x+3|=|(x-3)+6| \leq |x-3| + 6[/tex]

    By multiplying the statement above with |x-3| (which is positive) we got:

    [tex]|x+3||x-3| \leq (|x-3| +6)|x-3| [/tex]

    Now clearly [tex]6+|x-3|< 6 + \delta[/tex] and [tex]|x-3|<\delta[/tex]

    By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

    [tex](6+|x-3|)|x-3|< (6 + \delta)\delta[/tex]

    We can conclude that:

    [tex]|x+3||x-3| < (6 + \delta)\delta[/tex] if [tex]0 < |x-3| < \delta[/tex]

    Now isn't [tex]\epsilon = (6 + \delta)\delta[/tex] ??

    Why in my book they choose [tex](6 + \delta)\delta \leq \epsilon[/tex] ?

    And why they choose [tex]\delta \geq 1 [/tex] and with that restriction
    [tex](6 + \delta)\delta \leq 7\delta[/tex] and [tex]7\delta \leq \epsilon[/tex].

    Finally I don't understand why [tex]\delta = min(e/7,1)[/tex].

    Could somebody possibly explain?

    Thanks a lot.
     
  2. jcsd
  3. Nov 4, 2009 #2


    it is about setting a bound on delta so that you can make some sense out of |x+3||x-3| < epsilon. If |x-3| < delta and a constant (say, 5) were in place of |x+3| , then we could just easily say |x-3| < epsilon/5 right? yes. because 5 is not dependent on anything, epsilon/5 is just dependent on epsilon, so any possible epsilon1 you could think of can be expressed as epsilon/5. however, you cannot do this with epsilon/|x+3|, since now your inequality will be dependent on epsilon AND x.


    so let's pick a maximum size for delta, say 1. then |x-3| < 1 which makes -1 < x-3 < 1

    *note that we are allowed to do this, since delta is fixed in terms of epsilon. all we care about is that we are able to find some delta such that if 0<|x-a|< delta then |f(x) - L | < epsilon. we can make delta as small as we want, as long as it works. the limit will exist if there exists some "neighbourhood" around L so that all x in that area maps to some f(x) such that |f(x) - L | < epsilon.

    then 2 < x < 4 and then 5 < x+3 < 7 and that would mean that |x+3|< 7. now we can say that |x-3||x+3| < 7|x-3| which implies |x-3| < epsilon/7. So since we've set a limit to the possibilities of our delta, let's say that delta = min(1, epsilon/7) so we are sure that our delta will ALWAYS work.
     
    Last edited: Nov 4, 2009
  4. Nov 4, 2009 #3
    Thank you.

    But still I have many questions.

    First, why don't we set

    [tex]|x+3||x-3| < (6 + \delta)\delta = \epsilon [/tex]

    It is valid, right?

    [tex](\delta+3)^2-9 = \epsilon[/tex]

    [tex]\delta = 9 \pm \sqrt{9-\epsilon}[/tex]

    Hence we found delta so that the condition if 0<|x-a|< delta then |f(x) - L | < epsilon is satisfied.

    Also, I believe you got some errors in there

    5 < x+ 3 < 7 is not equivalent to |x+3| < 7 because |x+3|<7 <=> -10<x<4

    Also I don't know how did you conclude from:

    |x-3||x+3| < 7|x-3| that |x-3| < epsilon/7 :confused:
     
  5. Nov 4, 2009 #4
    1) well, generally, you want to find your delta in terms of epsilon, since delta depends on epsilon. only rarely will you be able to "find" delta such that epsilon can be expressed as delta.

    at any rate, it's never good to over-clutter a proof - the method of binding delta is far simpler and it is actually fairly standard.


    2) 5 < x+ 3 < 7 is not equivalent to |x+3| < 7 because |x+3|<7 <=> -10<x<4

    well no, x+3 is between 5 and 7, so |x+3| is still < 7




    3) |x-3||x+3| < 7|x-3| that |x-3| < epsilon/7 :confused:

    suppose 7|x-3| < epsilon, then |x-3| < epsilon/7.
    so |x-3||x+3|< epsilon , since |x-3||x+3| < 7|x-3|, and |x-3| < epsilon/7
     
  6. Nov 4, 2009 #5
    Yes |x+3| < 7 but it is not the same interval 5<x+3<7. Does it make any difference?

    Ok, somehow I figured 3) out .

    Because of |x+3|<7 then |x-3||x+3|<7|x-3| and because of |x-3| < delta then 7|x-3|<7delta so that |x-3||x+3|<7|x-3|<7delta

    So epsilon = 7 delta but in my book still epsilon >= 7 delta. Does that mean that we can choose some number less then epsilon/7 for delta? For example epsilon/14 ?

    And I still don't understand why we choose [tex]\delta = min(e/7,1)[/tex]. What if I choose e=14, will it still work?
     
  7. Nov 4, 2009 #6
    It's easiest to think of using the min function in this situation as allowing you to satisfy two inequalities at the same time. In this case, we required that |x-3| < 1 to simplify our analysis, and this lead us to |x+3| < 7, which allowed us to write |x-3||x+3| < 7|x-3|. Traditionally, we require |x-3| < delta which is in the definition of the limit, but in this case, note that we need |x-3| < e/7, yet we can't simply set delta = e/7 because we required |x-3| to be less than 1 AND less than e/7. The solution then is to choose delta to be the smaller of 1 and e/7 so that BOTH inequalities are satisfied, and this clearly works.

    If e = 14, min(1, 2) = 1 and we have |x-3||x+3| < 1*|x+3| < 7 < 14.
     
  8. Nov 4, 2009 #7
    1) we've already established that 5 < x+3 < 7 . the x in 5 < x+3 < 7 is the same x in |x+3|, so it is still |x+3| < 7

    2) not quite. epsilon is anything, delta is picked depending on epsilon - more specifically delta is epsilon divided by 7 (in this case). the idea is that you want to bind the function between (L- epsilon , L+ epsilon) - what values x should I pick to ensure this happens?
    well then, we are claiming that for any epsilon, |x+3||x-3| < epsilon. however, from the result before, we have concluded that x+3 is always less than 7, so |x+3||x-3| < 7|x-3|. now let's pick some epsilon for 7|x-3|, and so now we have 7|x-3| < epsilon => |x-3| < epsilon/7. Okay, what happened here? well, we are saying, "let's make 7|x-3| less than any epsilon I can possibly imagine" - but since 7|x-3| is less than any epsilon, and |x-3||x+3| is less than 7|x-3|, what does that leave us? We are left with the conclusion that |x-3||x+3| < epsilon; from our result, we can also deduce that |x-3| is less than epsilon/7. Well, |x-3| is exactly the same as the |x-3| from our |x-3| < delta inequality - but |x-3| < epsilon/7 is implied through-out our |x-3||x+3| premise. Therefore, if |x-3| < epsilon/7, then |x+3||x-3| < epsilon.


    we say that delta = min(1, epsilon/7) because we have said "all possible deltas are going to be 1 or less". since epsilon/7 can be anything, there is a possibility that delta = 1 will work better than epsilon/7. We want the smallest possible delta that can produce the x's that fit into our epsilon bound, so you may find that delta = 1 works better than epsilon/7 for some case.
     
  9. Nov 5, 2009 #8
    Ok, 7|x-3| < epsilon

    |x-3||x+3| < epsilon

    epsilon=15
    x=5
    15>14 from the first equation.

    15 > |5-3||5+3|

    15 > 16 (which is not true)

    So you can call 7|x-3| < epsilon1 but not epsilon. In fact I come to correct conclusion for 2). Why you said "not quite" ?

    Why this statement is not true:

    "
    Because of |x+3|<7 then |x-3||x+3|<7|x-3| and because of |x-3| < delta then 7|x-3|<7delta so that |x-3||x+3|<7|x-3|<7delta=epsilon" ??
     
  10. Nov 5, 2009 #9
    read this paper
     

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