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Epsilon and delta method of finding limits?

  1. Sep 22, 2011 #1
    first of all i dont know anything about this epsilon and delta method.explain this a bit.

    secondly i have been given a problem involving this method:
    f(x)=x^2

    given: limit x-->2 [x^2] = 4
    a) what is the value of x' such that f(x')= 4 + .01? find [itex]\delta[/itex]=x'-2
    b) what is the value of x[itex]_{2}[/itex] such that f(x[itex]_{2}[/itex])=4- .01? find [itex]\delta_{2}[/itex]=2-x[itex]_{2}[/itex]
    and finally from both of the values of delta prove that limit is equal to 4 at x->2

    i dont understand where to start.guide me
     
  2. jcsd
  3. Sep 22, 2011 #2
    Intuitively, when we speak of the limit of a function f at a point c, we mean that when x gets really close to c, then f(x) gets really close to this limit. Slightly more intuitively and more mathematical this is like saying "If you give me a positive number e," (take this to be epsilon, I'm not messing with Tex, and d will be delta) "I can give you a positive number d, such that if x is within d of c, that is |x-c|<d, then f(x) will be within e of the limit of f at c, that is |f(x)-Limit at c| < e." Now, for continuous functions, the limit at any point is the function evaluated at that point, ie the limit of f at c is f(c).

    OK, think about this for a bit, and see if it helps answer your question. Ask some more questions if you are still unsure about something.
     
  4. Sep 22, 2011 #3
    ok i see a bit now how epsilon and delta method. it is the nearly the same thing except that we dont pinpoint the limit rather say that if function is evaluted within this range of x then the output will be within some specified range of the limit.
    ok but what value do we give to epsilon and delta?
    please do this ....solve the first part for me and i will do the rest myself. that way i will gwt to know the procedure. my main problem is that i dont know where to start, whether to define epsilon first or what
    thanks
     
  5. Sep 22, 2011 #4
    OK, so you are getting it. The epsilon is given first, but you are NOT defining it. Someone gives you an e and you have to come up with a d such that if |x-c| < c then |f(x)-limit of f at c| < e. So, in some sense, the guy giving you the e is an adversary. That is, he says "OK, so you found a d for that e, but how about this e/2? can you find one for that?" In a general case, you will usually define e in terms of d.


    In this question:

    given: limit x-->2 [x^2] = 4
    a) what is the value of x' such that f(x')= 4 + .01? find δ=x'-2

    your prof has given you e, it is .01. He has said, "can you find a d such that if |x-2|<d then |x^2-4|<.01?" Before I solve this, I want you to do this:

    solve the inequality: x^2-4 < .01. Now, use that to find delta.
     
  6. Sep 22, 2011 #5

    symbolipoint

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    These might change everything for you:



     
    Last edited by a moderator: Sep 25, 2014
  7. Sep 23, 2011 #6
    ok....x^2-4<.01
    x^2<4.01
    which gives us that x<2.00249........
    i suppose now i put this x value in: |x-2|<d
    |2.0025-2|<d
    which dives us d>.0025
    so this means that whenever x will be .0025 greater than 2 i will get f(x) .01 greater than 4

    and for the b part the opposite will be true that is for .0025 smaller than 2 i get f(x) .01 smaller than 4
    now what?
    also if i am supposed to submit it as an assignment how should i write it...i mean the method?
     
  8. Sep 23, 2011 #7
    thanku very much for those links.....i understood them
    i just have one last doubt....in my problem i must replace epsilon by .01
    ryt?
     
    Last edited by a moderator: Sep 25, 2014
  9. Sep 23, 2011 #8

    SammyS

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    For part (b): "What is the value of x2 such that f(x2) = 4 - .01? find δ2=2-x2 ?"

    So (x2)2 = 4 - .01 . This gives δ2 ≈ 0.00250156, which is slightly larger than δ from part (a). ≈ 0.00249844.

    For ε = 0.1, a δ of 0.0025 is almost large enough. However, if x = 2.002499, then x2 = 2.002499^2 = 4.010002245001
    so that |f(2.002499) - 4 | > ε.
     
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