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

$\blacksquare$

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2 Answers 2

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You ask

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 $0$.

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  • $\begingroup$ "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? $\endgroup$ May 4, 2021 at 2:27
  • $\begingroup$ 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. $\endgroup$ May 4, 2021 at 11:57
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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 $\delta$).

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  • $\begingroup$ 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? $\endgroup$ May 4, 2021 at 1:15
  • $\begingroup$ ^^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 $\delta'$? $\endgroup$ May 4, 2021 at 1:15

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