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Help with proving limits using Epsilon-Delta definition

  1. Oct 3, 2004 #1
    I am having trouble proving the limits of quadratic functions such as the following. (I used "E" to represent epsilon and "d" for delta)

    lim [(x^2)+1]

    I found the limit, L, to equal 2 and have proceeded through the following steps:

    |f(x) - L| < E
    | [(x^2)+1] - 1| < E
    |[(x^2)-1]| < E
    |x+1||x-1| < E

    While I also know that 0 < |x-1| < d.

    My question is how do you find the numerical relationship between |x+1||x-1| and |x-1| so that I may find d in terms of E?

    (I was thinking of finding the bounded open interval in which |x+1||x-1| = |[(x^2)-1]| and substituting the greastest figure of the interval, which would be greater than |x+1|, in place of |x+1| so that I would have z|x-1|< E where z is an identified numerical value. However, in problems like the one above, |x+1||x-1| = |[(x^2)-1]| within a seemingly unbounded interval.)

    please help. your time and assistance is very much appreciated.
  2. jcsd
  3. Oct 3, 2004 #2
    Use the triangle inequality perhaps, |x + 1| = |x - 1 + 2| <= |x - 1| + 2 < d + 2.
  4. Oct 3, 2004 #3


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    Dearly Missed

    First of all, you have NOT found the limit, and certainly not shown that it is 2!
    What you have done, is to make a GUESS at the limit value!!
    (As it happens, you've made a correct guess, but that is irrelevant; it's still a guess).

    What you are to do now, is:
    1. Does my guess (2) satisfy the properties that a limit must have?
    2. Let [tex]|x-1|<\delta[/tex]
    3. Then, [tex]|x^{2}+1-2|=|x^{2}-1|=|x-1||x+1|[/tex]
    4. Now, by assumption, [tex]|x-1|<\delta[/tex]
    Let us make a further assumption, that [tex]\delta<1[/tex]
    Then, [tex]|x+1|<2 (\delta<1)[/tex]
    [tex]|x^{2}+1-2|<2\delta, (\delta<1)[/tex]
    5. Let [tex]\epsilon>0[/tex]
    If we are to have [tex]|x^{2}+1-2|<\epsilon[/tex] for all x satisfying [tex]|x-1|<\delta[/tex] it is sufficient if the following inequalities are simultaneously satisfied:

    Hence, setting [tex]\delta=min(\frac{\epsilon}{2},1)[/tex] suffices.

    That is, we were able to show that our guess (2) satisfy the properties a limit must have.
    Last edited: Oct 3, 2004
  5. Oct 4, 2004 #4
    By 'finding' 2 through the substitution of 1 in f(x)=[(x^2)+1] I meant that it was a possible limit whose validity must be varified through the E-d definition. Thank you for correcting me however, to avoid my own future confusion.

    Thanks for the helpful guidance and clarification.
  6. Oct 4, 2004 #5


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    Dearly Missed

    I was perhaps a bit too snappish on that point, don't bite me back, though..:wink:
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