Spivak Calculus, Ch. 5 Limits: Finding the smallest $b$ in $|x-3|<b$ such that $|x^2-9|=|x-3||x+3|<\epsilon$?

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 $|x^2-9|=|x-3||x+3|<\epsilon$?

Solution 1:

You have a lot of statements, so it is necessary to address each specifically.

$|x+3|=7$ only happens if $|x−3|=1$, the largest distance from $3$ that we allow, by assumption.

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 $4$."

Assuming this "worst case scenario", $|x−3|$ has to be relatively smaller so that the product $|x−3||x+3|$ is smaller than $\epsilon$.

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 $\epsilon < 7$.

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 $1$.

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 $|x^2 - 9| = 0.61 < 20$.

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 $|x−3|<\epsilon/7$.

Correct.

Assume that $\epsilon < 7$. Then, $|x−3|<\epsilon/7<1$. But then we can be more precise about how large $|x+3|$ is:

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.

How do I figure out what the smallest interval is for $x−3$ such that we have $|x^2−9|=|x−3||x+3|<\epsilon$?

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.

Solution 2:

Let $x = 3+h$ then the problem is to estimate $h$, now you need $$-\varepsilon< (3+h)^2 - 9<\varepsilon$$

which is same as saying $ 9-\varepsilon < (3+h)^2 < 9+\varepsilon$, Now we need to check cases if $\varepsilon < 9$ and othercases, using that you will be able to get the interval for $h$