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Math Help - For those that never learned well the epsilon-delta proofs

  1. #31
    Senior Member nycmath's Avatar
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    Re: For those that never learned well the epsilon-delta proofs

    I know that direct and indirect geometric proofs using the statement vs reason chart are not taught in most high schools across the USA. They cover the basics of proving two triangles are congruent but nothing beyond simple triangles. Teachers are forced to test prep their students. Teachers need to teach and spend less time preparing students for standardized exams. We need more education and less test prep.
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  2. #32
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    Re: For those that never learned well the epsilon-delta proofs

    I have a question regarding the more complicated proof in example 3 in the attachment. When you see that you can't define δ = ε/│x+5│and replace │x+5│ with M, why must│x+5│≤ M. Could you please explain this? Thanks so much for this post!
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  3. #33
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    Re: For those that never learned well the epsilon-delta proofs

    Quote Originally Posted by cdbowman42 View Post
    I have a question regarding the more complicated proof in example 3 in the attachment. When you see that you can't define δ = ε/│x+5│and replace │x+5│ with M, why must│x+5│≤ M. Could you please explain this? Thanks so much for this post!
    It's explained in the very next line. Basically, if x is reasonably close to 5, then we want a function of $\displaystyle \begin{align*} \epsilon \end{align*}$ that ends up being an upper bound for $\displaystyle \begin{align*} |x - 5| \end{align*}$. So say we want x to be 1 unit away from 5, then that means we have

    $\displaystyle \begin{align*} |x - 5| &< 1 \\ -1 < x - 5 &< 1 \\ 4 < x &< 6 \\ 9 < x + 5 &< 11 \\ | x + 5 | &< 11 \end{align*}$

    So that means as long as $\displaystyle \begin{align*} |x - 5| < 1 \end{align*}$ that means $\displaystyle \begin{align*} |x - 5| < \frac{\epsilon}{11} \end{align*}$, and thus an appropriate choice for $\displaystyle \begin{align*} \delta \end{align*}$ is $\displaystyle \begin{align*} \min \left\{ 1, \frac{\epsilon}{11} \right\} \end{align*}$.
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    Re: For those that never learned well the epsilon-delta proofs

    Ok, I'm just trying to fully grasp these concepts. What do you mean when you say a function of ε that ends up being an upper bound for │x-5│? I don't think I'm seeing the bigger picture here.
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    Re: For those that never learned well the epsilon-delta proofs

    That's the whole point of an $\displaystyle \begin{align*} \epsilon - \delta \end{align*}$ proof. To prove that $\displaystyle \begin{align*} \lim_{x \to c} f(x) = L \end{align*}$ you need to show that for all $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ there exists a $\displaystyle \begin{align*} \delta \end{align*}$ which is a function of $\displaystyle \begin{align*} \epsilon \end{align*}$ such that $\displaystyle \begin{align*} 0 < |x - c | < \delta \implies \left| f(x) - L \right| < \epsilon \end{align*}$.

    Loosely speaking, it's saying that when you set a tolerance for $\displaystyle \begin{align*} f(x) \end{align*}$ which we call $\displaystyle \begin{align*} \epsilon \end{align*}$, as long as you are sufficiently close to $\displaystyle \begin{align*} x = c \end{align*}$, then you are ensured that the distance between $\displaystyle \begin{align*} f(x) \end{align*}$ and the limiting value $\displaystyle \begin{align*} L \end{align*}$ is always less than your tolerance.

    So this "upper bound for |x - 5|" is the $\displaystyle \begin{align*} \delta \end{align*}$ function you are looking for in this case.
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