Consider the function $f(x)=x^2$.
In the text of chapter 5 of Spivak's Calculus, he goes through the following argument to show that $f(x)$ approaches the limit $9$ near $x=3$.
We want that $$|x^2-9|=|x-3||x+3|<\epsilon$$
Assume $|x-3|<1$.
$$\implies 2<x<4$$ $$\implies 5<x+3<7$$ $$\implies |x+3|<7$$
Therefore
$$|x^2-9|=|x-3||x+3|<7|x-3|<\epsilon$$
$$\implies |x-3|<\frac{\epsilon}{7}$$
provided $|x-3|<1$ as we assumed initially, ie $|x-3|<min(1,\frac{\epsilon}{7})$
I have a few observations about this argument:
- $|x+3|=7$ only happens if $|x-3|=1$, the largest distance from $3$ that we allow, by assumption.
- Assuming this "worst case scenario", $|x-3|$ has to be relatively smaller so that the product $|x-3||x+3|$ is smaller than $\epsilon$.
- In fact, depending on $\epsilon$, $|x-3|$ may have to be smaller than $1$. However, $\epsilon$ may also be so large that $|x-3|$ has to be smaller than a number larger than 1.
- All we know is that if $|x-3|<1$ then $|x+3|$ is definitely smaller than 7, and so to achieve a product smaller than $\epsilon$, we need $|x-3|<\frac{\epsilon}{7}$.
- Assume that $\epsilon<7$. Then, $|x-3|<\frac{\epsilon}{7}$<1. But then we can be more precise about how large $|x+3|$ is:
$$-\frac{\epsilon}{7}<x-3<\frac{\epsilon}{7}$$
$$6-\frac{\epsilon}{7}<x+3<6+\frac{\epsilon}{7}$$
$$|x+3|<6+\frac{\epsilon}{7}<7$$
$$|x^2-9|<(6+\frac{\epsilon}{7})|x-3|<\epsilon$$
$$|x-3|<\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
The interval allowed for $x$ near $3$ is now larger. But that means we can go back and calculate what this means for $|x+3|$ again:
$$-\frac{\epsilon}{6+\frac{\epsilon}{7}}<x-3<\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
$$6-\frac{\epsilon}{6+\frac{\epsilon}{7}}<x+3<6+\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
Note that $\frac{\epsilon}{6+\frac{\epsilon}{7}}>\frac{\epsilon}{7}$, so the interval for $|x+3|$ is now larger than before. And if we go back and recalculate the interval for $|x-3|$ it will now be smaller again.
My questions are
- Are my calculations correct? Ie, is it ok to do this thing where I narrow down the intervals one at a time?
- How do I figure out what the smallest interval is for $x-3$ such that we have $|x^2-9|=|x-3||x+3|<\epsilon$?
You have a lot of statements, so it is necessary to address each specifically.
While I understand what you intend to say here, it is not mathematically correct. To say $|x+3| = 7$ only happens if $|x-3| = 1$ would mean that the former condition can only be satisfied if the latter condition is true, and for no other values of $x$. However, $x = -10$ satisfies $|x+3| = 7$ but not $|x-3| = 1$. What you mean to say is that if $|x-3| \le 1$, then $|x+3| \le 7$, with equality occurring only when $x = 4$. The difference is that Spivak is saying, "if $P$ then $Q$," whereas what you wrote is "$Q$ only if $P$." These are not the same thing, in as much as the statement "a number divisible by $4$ is also divisible by $2$" is not the same as saying "a number is divisible by $2$ only if it is divisible by $4$."
The "worst case scenario" is that there is only one value of $x$ for which $|x-3| = 1$ and $|x+3| = 7$, namely $x = 4$. In this case, we can directly compute $|x-3||x+3| = 7$. Hence, if $\epsilon > 7$, a "radius" of $1$ around $x = 3$ is satisfactory to ensure $|x^2-9| < \epsilon$. However, this is neither the largest nor smallest choice; e.g., if $\epsilon = 7.00001$, then we can choose any radius $\delta$ satisfying $1 < \delta < 1.000001$ and this will result in a larger neighborhood for which $|x^2 - 9| < \epsilon$, and of course, we can choose a smaller neighborhood, say $|x-3| < 0.1$ and this will also work. It is only when $\epsilon < 7$ that the radius of $1$ will no longer work to ensure $|x^2 - 9| < \epsilon$, and in this case, Spivak seeks to quantify what radius will work. He shows that $\delta = \epsilon/7$ works when $\epsilon < 7$.
The first sentence is true, as I explained above. The second sentence is true but somewhat ambiguously stated. It is more precise to say that if $\epsilon > 7$ (i.e. "so large that"), then $|x-3|$ may be larger than $1$ while still satisfying $|x^2 - 9| < \epsilon$, but is not required to be. For instance, if $\epsilon = 20$, then the choice $x = 5$ gives $|x-3| = 2$ and $|x^2 - 9| = 16 < 20$, but the choice $x = 3.1$ gives $|x-3| = 0.1$ and $|x^2 - 9| = 0.61 < 20$.
Correct.
This iterative process is not necessary in this simple case. The precise maximal radius can be found because it is always corresponding to the right endpoint of the interval; i.e.,
$$\epsilon > |x^2 - 9| = x^2 - 9$$ whenever $x > 3$, hence $x = \sqrt{9 + \epsilon}$ is the right endpoint of the open interval satisfying the inequality, and so $$|x - 3| < \delta = -3 + \sqrt{9 + \epsilon} \tag{1}$$ is the $\delta$ of maximal size for any given $\epsilon > 0$. I leave it as an exercise to show that whenever Equation $(1)$ is true, $|x^2 - 9| < \epsilon$, for any $\epsilon > 0$ no matter how big or small.
There is no smallest interval. You are always free to pick an arbitrarily small interval. What you mean is the largest and therefore least restrictive interval, as I showed above.