Class example: limit of a function using definition

  1. I am having trouble understanding how to find the limit of a function (using the definition of a limit). I have a class example, and was wondering if anyone could walk me through the steps.

    1. The problem statement, all variables and given/known data
    Using the definition of the limit to show that limx→2(x2)=4
    f(x) = x2
    c=2
    L=4

    Given an arbitrary ε>0, take δ=min{1,ε/5}
    If x≠2 and |x-2|<δ then |x-2|<1 and |x-2|< ε/5
    |f(x)-L| = |x2-4| = |(x-2)(x+2)| = |x-2||x+2|
    |x-2|<1 => 1<x<3 => 3<x+2<5 => |x+2|<5
    |x-2||x+2| < (ε/5)(5) = ε so |f(x)-L|<ε


    2. Relevant equations
    We say that lim f(x)x→c=L if:
    [itex]\forall[/itex]ε>0 [itex]\exists[/itex]δ>0 [itex]\forall[/itex]x[itex]\in[/itex]dom f if x≠c and |x-c|<δ then |f(x)-ε|<L


    3. The attempt at a solution
    The biggest thing I am confused about is how the professor got δ? Did he have to do the later work first and then went back and plugged in the answer he got?

    Also, in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε. Why is this? I understand that we can rearrange the equation, but then doesn't this mess up the absolute value signs?
     
  2. jcsd
  3. SammyS

    SammyS 8,039
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    "The biggest thing I am confused about is how the professor got δ? Did he have to do the later work first and then went back and plugged in the answer he got?"
    Your professor likely did some scratch work, starting with |x2-4|<ε, and then getting his result for δ. ​

    "in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε"

    It should be |f(x)-L|<ε in the definition.​
     
  4. Dick

    Dick 25,735
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    |f(x)-epsilon|<L is a typo. |f(x)-L|<epsilon is the correct form. And yes, the professor figured out a delta using the later work and then went back and plugged it in.
     
  5. Another question, is there more than one δ that will prove this?
    Say, Given an arbitrary ε>0, take δ=min{2,ε/6}
    If x≠2 and |x-2|<δ, then |x-2|<2 and |x-2|<ε/6
    |f(x)-L| = |x2-4| = |(x+2)(x-2)| = |x+2||x-2|
    |x-2|<2 => -2<x-2<2 => 0<x<4 => 2<x+2<6 => |x+2|<6
    |x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε
     
  6. Dick

    Dick 25,735
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    Sure, that choice works just as well.
     
  7. SammyS

    SammyS 8,039
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    Yes, there are many ways to come up with δ .
     
  8. Why is it that I must say d=min{1,ε/5}? Would it also work if I said that δ=1,ε/5. I'm a bit fuzzy on how the "min" makes this true, or the absence of "min" makes it false.
     
  9. Last edited: Nov 13, 2011
  10. SammyS

    SammyS 8,039
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    If ε > 5, then if you say that δ > ε/5, the proof won't work.

    Added in Edit:
    Let's say ε = 10.

    Then the claim would be that δ = 2 will satisfy the definition.
    But if x=3.9, then f(3.99)=15.21, so |f(3.99)-2| = 13.21 > 10
     
    Last edited: Nov 13, 2011
  11. So is it standard procedure to always take δ=min if there is more than one condition? Will it ever be wrong for me to make δ=min ?
     
  12. Dick

    Dick 25,735
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    In the proof you used that d<=1 AND d<=epsilon/5. min(1,epsilon/5) is less than or equal to both of them. d=1 doesn't work if you pick a small epsilon. d=epsilon/5 doesn't work if you pick a large epsilon. Try it.
     
  13. Thank you! This makes so much more sense now!
     
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