Help with proving limits using Epsilon-Delta definition

  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]
    x->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. Use the triangle inequality perhaps, |x + 1| = |x - 1 + 2| <= |x - 1| + 2 < d + 2.
     
  4. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    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]
    And:
    [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:
    [tex]2\delta<\epsilon[/tex]
    [tex]\delta<1[/tex]

    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. 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. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

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