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Understanding epsilon-delta def of limits

  1. Jun 21, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm having trouble conceptually understanding the epsilon-delta definition of limits. How do you use it to disprove a limit? For example, how would you use it to show that [tex]\lim_{x \to 1} x^2 \neq 2[/tex]?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 21, 2008 #2
    imagine you've got the function x plotted in a graph(you can actually draw it in a piece of paper). and you wanna show that [tex]
    \lim_{x \to 1} x \neq 2[/tex]. First, set an interval ]1-delta,1+delta[ in x axes, see the image of the 1+delta; epsilon will be=|f(1+delta)-f(1)|. (in general, it will be the max{|f(1+delta)-f(1)|,|f(1-delta)-f(1)|}, in this case, the function is similar to the right and left of f(1), so, only one part is needed because both are equal).
    Now, you see that if the delta is very big, say 4, then epsilon=4 rigth?! so, (the second part of the condition, the epsilon part) is |2-f(1)|=2-1=1<epsilon: the condition is verified.(2 because you are testing the condition on the point 2 you are asking: "is it really true that the condition is verified to EVERY delta??????")

    Note however that if you put delta lower, say 0.5, epsilon is 0.5 and |2-f(1)|=|2-1|>epsilon:. Exists a delta that don't verify the condition, that implies that isn't true that for every delta, exists..bla bla bla...so, the limit isn't 2.

    Hope this helps solve your problem...
    and hope i didn't make any mistake...^_^
     
    Last edited: Jun 21, 2008
  4. Jun 21, 2008 #3
    Perhaps in this way
    can Be clear idea

    [​IMG][/url][/IMG]
     
  5. Jun 21, 2008 #4
    but how do you put that into an epsilon-delta proof?
     
  6. Jun 22, 2008 #5
    [​IMG][/url][/IMG]
    [​IMG][/url][/IMG]

    No requirement that the value of f(x0)

    The smaller values for delta & epsilon Leads to ........what?
     
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