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Homework Help: Showing by definition that limits exist

  1. Sep 28, 2004 #1
    I am having trouble with showing by definition that limits exist. For example limit as x goes to 3 for 5(x^2)-17

    3-D < x < 3+D

    28-E < 5x^2 - 17 < 28+E
    45-E < 5x^2 < 45+E
    9-E/5 < x^2 < 9+E/5
    sqrRoot(9-E/5) < x < sqrRoot(9+E/5)

    so does it mean that 3-D=sqrRoot(9-E/5) ?
    then D=3+-sqrRoot(9-E/5)? If it is correct, can you do it without knowing the limit?

    In linear equation you can solve it without even knowing the limit:

    lim x goes to 3 for 5x - 17

    3-D < x < 3+D

    L=limit (5*3-17=-2)

    L-E < 5x - 17 < L+E
    L+17-E < 5x < L+17+E
    (L+17)/5 - E/5 < x < (L+17)/5 + E/5


    3- E/5 < x < 3+ E/5

    So D=E/5
  2. jcsd
  3. Sep 28, 2004 #2


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    Dearly Missed

    1. When doing limits exercises, introduce "epsilon" at the latest possible stage in your actual calculations!
    2. Let's take your exercise in detail:
    Let [tex]|x-3|<\delta[/tex]:
    Then, we have, guessing at 28 as our limit:
    Assume [tex]\delta<1[/tex]
    By choosing [tex]\delta=minimum(1,\frac{\epsilon}{20})[/tex]
    we have bounded our error within a margin of [tex]\epsilon[/tex]
  4. Sep 28, 2004 #3
    how do you get [tex]5|x-3||x+3|<5\delta{4}=20\delta[/tex] ?
  5. Sep 28, 2004 #4
    This is as far as I get.......

    |5x^2-17 - 5x0^2+17|<E

    and |x-3|<D
    then D=E/(5|x+3|)

    How do you get 20?
  6. Sep 29, 2004 #5


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    Dearly Missed

    By assuming D<1, |x+3|<4.
    As I've said, do not introduce E in your calculations before you have to.
  7. Sep 29, 2004 #6
    I see, but this does not apply to every function does it?

    lim as x goes to 0 for 1/x^2


    |1/x^2 - 1/x0^2| =
    |(x0^2-x^2)/(x^2)(x0^2)| =
    |(x^2-x0^2)/(-x0^2)(x^2)| =

    and since x0=0 then D=0 ???
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