Necessity of selecting value for $\delta$ in $\epsilon$-$\delta$ proofs

Question

I've been reviewing the epsilon-delta definition of a limit before beginning Analysis 1 and am slightly confused about the problems where $\delta$ is selected from a minimum of two values.

For example, a problem in my textbook asks to prove $$\lim_{x \to 3} x^2 = 9$$

using epsilon-delta. The process is explained as follows:

$$|x^2-9|<\epsilon$$

$$|x+3||x-3|<\epsilon$$

$$\text{Let } \delta = 1, \text{so } |x-3|<1$$

$$-1<x-3<1$$

$$5<x+3<7$$

$$\therefore |x-3|<\frac{\epsilon}{7}$$

$$\delta = \min \{ 1, \frac{\epsilon}{7} \}$$

I'm confused as to why we don't leave delta as delta, and simply solve for delta in terms of epsilon (i.e. $\delta = \sqrt{\epsilon+9}-3$). Are we able to simply pick a bound for $\delta$ because epsilon-delta proofs only require that there exists a positive delta that will work, and that there an infinite number of positive deltas that exist in $\left[0,1\right)$, for example? Otherwise, the proof seems incomplete because it doesn't take into account any $\delta>1$.

Answer 1

Yes, the definition only requires that, for every $\varepsilon>0$, there is some $\delta>0$ such that$$|x-a|<\delta\implies\bigl|f(x)-f(a)\bigr|<\varepsilon.$$So, the fact that, in that example, the $\delta$ is always smaller than or equal to $1$ is irrelevant.

And, yes, in some cases we can get a $\delta$ that works as a function of $\varepsilon$, but in most cases we cannot. Your example ($\delta=\sqrt{\varepsilon+9}-3$) is indeed correct, but try to do the same thing with $\lim_{x\to1}x^3+x^2+x=3$.

Answer 2

The answers given so far all focus on the narrow question of the choice of $\delta$, but don't address what appear, to me, to be more general confusions on the part of OP about $\epsilon$-$\delta$ proofs. If OP is going to take analysis, it would be good to clear up those confusions.

OP writes that "the proof seems incomplete." But no proof has been given. Does OP think that the steps listed under "The process is explained as follows" are a proof? They are not. They are simply the scratch work to find a good choice for $\delta$. The proof must be an essay establishing the truth of the statement: For every $\epsilon>0$ there is some $\delta>0$ such that for every number $x$, if $0 < |x-3| < \delta$ then $|x^2-9|<\epsilon$. Using the choice $\delta = \min\{1,\epsilon/7\}$, the proof would be written like this:

Suppose $\epsilon > 0$. Let $\delta = \min\{1, \epsilon/7\}$. Then $\delta \le 1$, $\delta \le \epsilon/7$, and $\delta > 0$. Suppose $0 < |x-3|<\delta$. Then $|x-3| < 1$ and $|x-3|<\epsilon/7$. Since $|x-3|<1$, $-1<x-3<1$, so $5 < x+3 < 7$ and therefore $|x+3|<7$. Thus $$|x^2-9| = |x+3|\cdot |x-3| < 7 \cdot \epsilon/7 = \epsilon,$$ which completes the proof.

OP says "solving for $\delta$" gives the result $\delta = \sqrt{\epsilon+9}-3$. I don't know what this means; there is no equation you can solve for $\delta$. If you want to know if this choice of $\delta$ is acceptable, you have to try to write the proof. Let's try it out:

Suppose $\epsilon > 0$. Let $\delta = \sqrt{\epsilon+9}-3 > 0$. Suppose $0 < |x-3| < \delta$. Then $-\delta < x-3 < \delta$, so $3-\delta < x < 3+\delta$. If $\delta \le 3$, then we have $$0 \le 3-\delta < x < 3+\delta,$$ and we can use the fact that the squaring function is increasing on $[0, \infty)$ to conclude that $$(3-\delta)^2 < x^2 < (3+\delta)^2.$$ By the choice of $\delta$, we have $(3+\delta)^2 = 9+\epsilon$. (Is this the equation OP solved to get $\delta$? If so, it is relevant to the proof, but it doesn't solve the whole problem.) So we have $$(3-\delta)^2 < x^2 < 9+\epsilon.$$ If we could now prove that $(3-\delta)^2 \ge 9 - \epsilon$, then we would have $$9-\epsilon < x^2 < 9+\epsilon,$$ from which we could conclude that $|x^2-9|<\epsilon$, as required to complete the proof.

So there are still two gaps in the proof: We must show that if $\delta = \sqrt{\epsilon+9}-3$ and $\delta \le 3$ then $(3-\delta)^2 \ge 9-\epsilon$, and we must give a separate argument to cover the case $\delta > 3$. Now, those gaps can be filled in, so this choice of $\delta$ can be used to give a correct proof. But it is already clear that this proof is turning out to be harder than the previous one.

The main point is this: In order to tell if a proposed choice of $\delta$ is acceptable, you have to write the proof. Any choice that allows you to complete the proof is acceptable. A good choice is one that leads to a simple proof. The purpose of the exercise is not to "solve for $\delta$"; the purpose is to write a proof.

Answer 3

$\lim x^{2}=9$

For a given $\in>0$, you need to find $\delta>0$ such that $0<|x-3|<\delta \Rightarrow\left|x^{2}-9\right|<\epsilon$

One approach is to just solve inequality $\left|x^{2}-9\right|<\epsilon$ as folows $\left|x^{2}-9\right|<\epsilon \Rightarrow \quad 9-\epsilon<x^{2}<9+\epsilon$ $\Rightarrow \sqrt{9-\epsilon}<x<\sqrt{9+\epsilon}$

Now you need a $\delta$ for which $(3-\delta, 3+\delta)$ sits inside the interval from $\quad \sqrt{9-\epsilon}$ and $\quad \sqrt{9+\epsilon}$

i.e. $\quad \sqrt{9-\epsilon}<3-\delta$ and $3+\delta<\sqrt{9+\epsilon}$ $$ \Rightarrow \quad \delta<3-\sqrt{9-\epsilon} \text { and } \delta<\sqrt{9+\epsilon}-3 $$ so biggest value of $\delta$ that would work is $$ \delta=\min \{3-\sqrt{9-\epsilon}, \sqrt{9+\epsilon}-3\} $$ But if function is a polynomial as in this case then another approach is to look at $\frac{f(x)-f(a) \text { . }}{x-a}$

If you can guarantee that $\quad\left|\frac{f(x)-f(a)}{x-a}\right| \leq c \quad$ it's safe to take $\delta=\epsilon / c$. because then $|x-a|<\delta \Rightarrow|f(x)-L|=|f(x)-f(a)|$ $$ =|x-a| \cdot\left|\frac{f(x)-f(a)}{x-a}\right| <\delta.c = \epsilon $$ But in practice one usually can't find a $c$ without assuming $x$ is bounded. But this is okay we can always take $\delta$ to be smaller of the two numbers In this case 1 or $\epsilon / 7$. This appreach always works for polynomials and often works for rational funetions.