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!
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.
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 $\displaystyle \begin{align*} \min \left\{ 1, \frac{\epsilon}{11} \right\} \end{align*}$.
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.
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.
I am on my IPad so going back and forth is almost impossible, but I think there are two issues that you are missing. First, if the absolute value is less than one fifth, it is necessarily less than one third. So one fifth works, but it is unnecessarily restrictive. Second and more important, 0 < a < b < c is equivalent to 0 < 1/c < 1/b < 1/a. As I say, it is hard for me to figure out what you are thinking and how it relates to an attachment on a different page without being on my PC, but as far as I can tell under not good conditions, there is no error in the example.
If you explain what you think the error is, I'll look again tomorrow on the PC.