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Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01 (f(x) = x^2 -3)

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01

    Limit when x approaches 2 of x^2 -3

    2. Relevant equations
    0 < |x-c| < δ
    0 < |f(x) - L| < ε

    3. The attempt at a solution
    function is continuous at D = ℝ so limit is f(2) = 1

    0 < |x-2| < δ and

    0 < |x^2 -3 -1| < 0.01 ⇔ 0 < |x^2 -4| <0.01 ⇔ 0 < |x + 2|*|x-2| < 0.01 which is good because:
    0 < |x-2| = 0.01 / |x+2|

    I've seen the solution and i see that i'm supposed to assume a range for x (like (1,3) ). I can imagine that because the function isn't linear a range has to be assumed.

    They say that assuming this range gives δ = 0.01 / 5 = 0.002 witch seems to be the smallest of 0.01 / 3 and 0.01 / 5. That makes sense to me.

    But why this chosen range? Why does this range apply to ε = 0.01? Choosing a different range gives a different δ.

    Who can help me with this i'm really trying to understand this.
     
  2. jcsd
  3. Oct 14, 2012 #2

    tiny-tim

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    Science Advisor
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    hi ialink! :smile:
    exactly! :smile:

    δ depends on ε

    (we could write it δ(ε) )

    these proofs all involve showing that whatever ε we choose, we can always find a δ :wink:

    (usually δ gets smaller and smaller, just like ε)
     
  4. Oct 14, 2012 #3
    Hey Tiny Tim

    If i Would express δ(ε) I'd say δ(ε) = ε / |x+2|. Calculating δ with this function is possible for D = ℝ (for both ε and x) so a logical conclusion is that the limit exists that's what you mean right?

    but ε = 0.01 is given to find the appropriate δ. Something has to be assumed for x in δ(ε) = ε / |x+2|. They assume (1,3) and find δ = 1/5*ε or δ = 1/3*ε and therefore conclude that δ = 1/5 * 0.01 = 0.002. That i don't get. What's the relation between the chosen range (1,3) an ε = 0.01? Why is that valid?
     
    Last edited: Oct 14, 2012
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