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Math Help - Epsilondelta Proof

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    Epsilondelta Proof

    Based off of the pdf document "those who never learned to do epsilondelta." How do you solve for |x-0| in the following \lim_{x\to 0}\frac{1}{x^2-1}=1. This is what I currently have: \frac{|x^2|}{|x^2-1|}\leq \epsilon
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    Re: Epsilondelta Proof

    Quote Originally Posted by brucewayne View Post
    \lim_{x\to 0}\frac{1}{x^2-1}=1.
    That's not true, so no correct proof can every demonstrate it.

    \lim_{x\to 0}\frac{1}{x^2-1}=-1.
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    Re: Epsilondelta Proof

    Quote Originally Posted by brucewayne View Post
    \lim_{x\to 0}\frac{1}{x^2-1}=1.[/TEX]
    As johnsomeone pointed out, the limit is -1. Use that \left |\dfrac{1}{x^2-1}+1\right |=x^2\dfrac{1}{|x^2-1|} and \dfrac{1}{|x^2-1|}\leq \dfrac{4}{3} in [-1/2,1/2].
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    Re: Epsilondelta Proof

    Yes, you are correct, it should equal -1.
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    Re: Epsilondelta Proof

    also, just to point out: there is nothing special about 1/2, we can pick ANY a with 0 < a < 1, in which case:

    |x| < a \implies \frac{1}{|x^2 - 1|} < \frac{1}{1-a^2}

    if we call this number L, what can we say about x for which:

    |x| < \min\left\{a,\sqrt{\frac{\epsilon}{L}}\right\} ?

    (choosing a close to 1 will make L very large, which will make our choice of delta fairly small. that's ok. choosing a close to 0 will make L a bit larger than 1, and whether or not a is smaller than √(ε/L) depends on how small ε is. choosing a = 1/2 represents a "middle ground" where the numbers are easy to calculate with, in this case, and we can focus on ε, rather than the expressions involving a, which can be distracting. but a = 1/3, or a = 1/4, for example, would work just as well).

    slight quibble:

    since we are finding a LIMIT, rather than checking for continuity, we want a DELETED neighborhood of 0, that is: 0 < |x| < δ. in this example, it doesn't matter, the function being checked:

    f(x) = 1/(x2 - 1) is continuous (but unbounded) on (-1,1), so on any sub-interval.

    but, in general, f(0) doesn't have to exist for the limit of f(x) as x approaches 0 to exist, and even if f(0) does exist, it may not equal the limit. that is: for CONTINUOUS functions, you can find the limit:

    \lim_{x \to 0} f(x)

    by just finding f(0), but if one does not know whether a function f is continuous at 0 or not, to find the limit you don't let x "go to 0", it only gets "near it".

    the classic example of this is:

    \lim_{x \to 0} \frac{x}{x} = 1.

    note we cannot consider values in any interval containing 0, since x/x is undefined at 0. if we define:

    g(x) = x/x , x ≠ 0
    g(0) = 5

    g is a perfectly good function, and the limit is still 1 at 0, but it isn't true that for ALL x with |x| < δ, |g(x) - 1| < ε, there is one exception: x = 0, in which case
    |g(x) - 1| = 4 (and so if ε < 4, we're toast).
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