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Math Help - Epsilon Delta Proof with Infinity

  1. #1
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    Epsilon Delta Proof with Infinity

    Hi,

    I have a question about an Epsilon Delta proof in the form of f(x) = L, where x goes to infinity and L is 1/3.

    (x^2 + x - 1) / (3*x^2 - 1) = 1/3 (where x goes to infinity)


    For the question I did the following things:

    I wrote the whole expression (A) in the following way:

    A - (1/3) < Epsilon



    The function did not get simplified nicely at the end, and I had to to use the square root formula to get x at the end:

    I got this as inequality for x:

    x > (1 + sqrt(1 - 8Epsilon + 12Epsilon^2)) / 6Epsilon

    From that point I didn't know how to continue. I will be more than glad, if you could give me an idea.
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  2. #2
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    Quote Originally Posted by kaka87 View Post
    Hi,

    I have a question about an Epsilon Delta proof in the form of f(x) = L, where x goes to infinity and L is 1/3.

    (x^2 + x - 1) / (3*x^2 - 1) = 1/3 (where x goes to infinity)


    For the question I did the following things:

    I wrote the whole expression (A) in the following way:

    A - (1/3) < Epsilon



    The function did not get simplified nicely at the end, and I had to to use the square root formula to get x at the end:

    I got this as inequality for x:

    x > (1 + sqrt(1 - 8Epsilon + 12Epsilon^2)) / 6Epsilon

    From that point I didn't know how to continue. I will be more than glad, if you could give me an idea.
    Just a simple idea.
    Let f(x):=\frac{x^{2}+x-1}{3x^{2}-1} for x\geq\sqrt{1/3} (we may replace \sqrt{1/3} with a larger one since we take limit as x\to\infty).
    Define g(x):=f(1/x)=\frac{x^{2}-x-1}{x^{2}-3} for x\geq\sqrt{3}.
    Now, (although it is obvious) show that \lim_{x\to0^{+}}g(x)=1/3, which is equivalent to showing \lim_{x\to+\infty}f(x)=1/3
    Last edited by bkarpuz; September 20th 2009 at 06:01 AM.
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  3. #3
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    Thank you very much However, could you please describe it more clearly? I am new to epsilon delta proofs, so I don't know the intermediate steps very well (Considering the last part of your idea).
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by kaka87 View Post
    Thank you very much However, could you please describe it more clearly? I am new to epsilon delta proofs, so I don't know the intermediate steps very well (Considering the last part of your idea).
    Clearly, you can see that the g is increasing in a neighborhood of 0 (you can see it by differentiating) and g(0)=1/3. We have to show that for every \varepsilon>0 there exists \delta>0 such that |x|\leq\delta implies |h(x)|\leq\varepsilon, where h(x):=g(x)-1/3=\frac{x(2x-3)}{3(x^{2}-3)} for x\in\mathbb{R}. Note also that since g is increasing, then h is also increasing and h(0)=0, which implies that h is positive in a left neighborhood of 0 while it is positive in right neighborhood.
    Now consider the case, x\leq0 but arbitrarily closer to 0. Then, for every \varepsilon>0, we have to find \delta_{1}>0 such that -\delta_{1}\leq x\leq0 implies |v(x)|=-v(x)\leq\varepsilon.
    By considering the increasing nature of v, we solve -v(-\delta_{1})=\varepsilon and get \delta_{1}(\varepsilon):=\frac{3(-1+\sqrt{1+8\varepsilon+12\varepsilon^{2}})}{2(3\va  repsilon+2)}>0 (we picked the positive root of the quadratic equation).
    Now let x\geq0 but again closer to 0.
    Then, for every \varepsilon>0, we have to find \delta_{2}>0 such that 0\leq x\leq\delta_{2} implies |v(x)|=v(x)\leq\varepsilon.
    In this case, we solve v(\delta_{2})=\varepsilon and get \delta_{2}(\varepsilon):=\frac{3(-1+\sqrt{1-8\varepsilon+12\varepsilon^{2}})}{2(3\varepsilon-2)}>0 (we picked the smaller root in a very small neighborhood of 0).
    We therefore for every \varepsilon>0 have |v(x)|\leq\varepsilon whenever |x|\leq\delta(\varepsilon), where \delta(\varepsilon):=\min\{\delta_{1}(\varepsilon)  ,\delta_{2}(\varepsilon)\}=\delta_{1}(\varepsilon)  >0 for \varepsilon>0.
    This completes the proof.
    But I have to say that to learn \varepsilon-\delta technique this example is not a good start...
    Last edited by bkarpuz; September 20th 2009 at 07:34 AM.
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