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Homework Help: Definition of a Limit

  1. Nov 14, 2013 #1
    The problem statement, all variables and given/known data

    I know what a limit is and I understand the idea behind it, but I am misunderstanding something in the formal definition of a limit.

    Let f(x) be defined on an open interval about x0, except possibly at x itself. We say that the limit of f(x) as x approaches x0 is the number L and write

    limx → x0 f(x) = L

    if, for every number ε > 0 there exists a corresponding number δ > 0 such that for all x,
    0 < abs(x – x0) < δ
    abs(f(x) – L) < ε

    The attempt at a solution

    For example, take f(x) = x2

    I could, incorrectly, assert that limx → 2 f(x2) = 1

    Now, δ > abs(x - 2) > 0
    and ε > abs(f(x) - 1)

    1) this holds true for all values of x (except x = x0 = 2), and,
    2) for every possible value of epsilon some value of x which satisfies the above statement for both ε and δ exists,
    1 is the limit of x2 as x approaches 2.

    As an example, I pick x = 3. So,
    δ must be > 3
    ε must be > 8

    I can continue and choose arbitrary values of x, none of which seem to be a problem.

    Likewise, for any value of epsilon I can think of, I can find an value of x which satisfies the inequality ε > abs(f(x) - 1) which also satisfies δ > abs(x - 2) > 0.
    ex: when ε > .01 I could have x = .999. This x also satisfies δ > abs(x - 2) > 0
    (as far I can tell, it's impossible not to satisfy gamma, as long as x isn't equal to x0)

    By that reasoning, the claim I made above is true and limx → 2 f(x2) = 1

    By the reasoning I'm using with the formal definition, I can state that any value of L is the limit for any function at an arbitrary x0 (given that the function has values infinity close to L at some point in time). This is clearly incorrect. Would anyone please point out the error in my application of the definition of a limit?

    Thank you.
  2. jcsd
  3. Nov 14, 2013 #2


    Staff: Mentor

    Your error is that someone else chooses ##\epsilon##, and from that you have to find a δ > 0 so that, when |x - 2| < δ, |f(x) - L| < ##\epsilon##.

    Using your example, where you postulate that ##\lim_{x \to 2}x^2 = 1##, I will say that ##\epsilon = .01##. You now have to find a pos. number δ such that for each x ##\in## (2 - δ, 2 + δ), then x2 ##\in## (.99, 1.01).

    Sketch a graph of the function y = x2, and you'll see that this can't happen.
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