# Confused on how we choose $\delta$ in $\epsilon, \delta$-proofs.

I am trying to prove the following statement rigorously using an $$\epsilon,\delta$$-proof.

Prove if $$b \in \mathbb{R}^+$$ and $$\lim_{x\to 0}\frac{f(x)}{x} = L$$, then $$\lim_{x \to 0} \frac{bx}{x}=bL$$.

There are a couple posts on Math StackExchange related to this problem. Specifically:

I understand how to manipulate the symbols to prove the statement without using the $$\epsilon,\delta$$ concept (i.e. noting that $$x \to 0$$ is the same as $$bx \to 0$$ $$\forall b \in \mathbb{R}^+$$). However, both these answers contain proofs with a step I do not understand (detailed below).

I have written the proof below in a more pedantic manner, attempting to explain why each step is happening, and highlighting my question inline.

Proof: Let $$\lim_{x \to 0}\frac{f(x)}{x} = L$$. Then, by definition, $$\forall \epsilon, \exists \delta'$$ s.t. $$\forall x, \lvert{x}\rvert < \delta' \implies \lvert{\frac{f(x)}{x} - L}\rvert < \epsilon$$. Since this is true for any epsilon, and because $$b \in \mathbb{R}^+$$, we know $$\frac{\epsilon}{\lvert{b}\rvert}$$ is well defined. Thus, $$\lvert{\frac{f(x)}{x} - L}\rvert < \frac{\epsilon}{\lvert{b}\rvert}$$.

Now, given any $$\epsilon > 0$$, we choose $$\delta = \frac{\delta'}{\lvert{b}\rvert}$$. Then $$\lvert{x}\rvert < \delta \implies \lvert{x}\rvert < \frac{\delta'}{\lvert{b}\rvert}$$. (My question: how is this justfied? How can we simply pick this new $$\delta$$? It seems like magic to me. What if this $$\delta$$ is larger than the $$\delta'$$ we used to deduce the limit? After all, we could choose $$b$$ so that $$-1 < b < 1$$, right? Or are we saying that this is the condition whereby the statement is true, and must be able to find such a $$\delta$$?)

Then $$\lvert{x}\rvert < \frac{\delta'}{\lvert{b}\rvert} \implies \lvert{bx}\rvert < \delta' \implies \lvert{\frac{f(bx)}{bx} - L}\rvert < \frac{\epsilon}{\lvert{b}\rvert} \implies \lvert{b}\rvert\lvert{\frac{f(bx)}{bx} - L}\rvert < \epsilon \implies \lvert{\frac{f(bx)}{x} - bL}\rvert < \epsilon$$.

Thus, by definition, $$\lim_{x \to 0}\frac{bx}{x} = bL$$.

$$\blacksquare$$

How can we simply pick this new Ξ΄? It seems like magic to me. What if this Ξ΄ is larger than the Ξ΄β² we used to deduce the limit?

In this adversarial game against an opponent who provides $$\epsilon$$ and challenges you to find a $$\delta$$ you are free to use any method that works.

If you have access to magic, go for it. If not, the usual way to find a $$\delta$$ is to manipulate the $$\epsilon$$ inequality to see what it tells you about the independent variable - in this example, how close $$x$$ must be to $$0$$. Then you write the inequality logic in the reverse order, starting from an inequality using $$\delta$$ to end with the $$\epsilon$$ one you want. That's just what you did.

Any working $$\delta$$ is fine. In fact, there will always be many such, because $$\delta$$ tells you how close $$x$$ must be to $$0$$ (in your example). If a particular $$\delta$$, say $$0.02$$ works then so will any smaller value, so you could choose $$\delta = 0.01$$ or $$\delta = 0.00001$$ if you wanted to.

In your example, suppose $$b = 1/2$$. Then your argument shows that you can choose $$\delta = 2\delta^\prime$$, which is indeed larger than $$\delta^\prime$$. That's OK. With that value of $$b$$ you are approaching the $$0$$ limit "faster" so you have more leeway when you determine how near $$x$$ must be to $$0$$.

• "If you have access to magic, go for it" hah! ^_^ I think your post and Arthur both really helped. The crux of my problem is that I keep thinking that $\delta'$ is being used to somehow *derive& $\delta$...but based on both of your posts, that is incorrect. In fact, I am plucking a $\delta$ out of the void of non-empty sets and then using the fact that I have a useful statement about $\delta'$ to build my brand new $\delta$. Does that make sense? May 4, 2021 at 2:27
• I think you use $\delta^\prime$ to guide your search for $\delta$, so the process is something creative in between deriving and making a blind pick from the void. May 4, 2021 at 11:57

It's a bit tricky to answer questions like 'How can we simply pick this new $$\delta$$?' Choosing an element in a non-empty set is just something you can always do in standard mathematics.

From the assumption that $$\lim_{x\to 0} \frac{f(x)}{x} = L$$ we know that, for any $$\epsilon>0$$, the set of numbers satisfying

$$\left|\frac{f(x)}{x}-L<\frac{\epsilon}{|b|}\right| \tag{1}$$

is non-empty. Therefore we can pick one $$\delta'$$. Once we've picked it we can do what we like with it, in particular we can define $$\delta = \frac{\delta'}{|b|}$$. Then, by exactly the argument you described, we have

$$|x|<\delta \implies \left|\frac{f(bx)}{x}-bL\right|<\epsilon \tag{2}$$

which proves that $$\lim_{x\to 0} \frac{f(bx)}{x} = bL$$. You rightfully point out that $$\delta$$ may be larger than $$\delta'$$. This is true but irrelevant. The only fact about $$\delta'$$ that your proof of (2) uses is that it satisfies $$(1)$$. You don't need to know anything else about it (in particular, not that it is larger that $$\delta$$).

• I think I completely get it!! My confusion is that we were using the $\delta'$ in order to somehow derive $\delta$...but that's not it at all, is it? We are just saying: (1) that this limit exists is expressed by this mathematical statement, and then (2) suppose we have pick other delta in the universe of non-empty sets (in this case, a set related to our $\delta,\epsilon$ statement), regardless of the fact that $\delta'$ exists. Then later, we say something like..."wow! look at that. If we use $\delta'$, then we can deduce our $\epsilon$ statement." Is that correct? May 4, 2021 at 1:15
• ^^In fact, we could have started the argument with "Suppose $\exists \delta$" and then later stated our fact about $\delta'$, and while it would have been a weird proof to read, it would be logically correct, since picking $\delta$ does not depend on the existence of $\delta'$ until we want to the inequality presented by $\delta'$? May 4, 2021 at 1:15