A seemingly naive question on continuous functions

Question

This question was motivated by the argument in Possible progress on the Riemann hypothesis? where the Riemann hypothesis is purportedly proved.

The crux of the OP's argument claiming to prove the RH is that, IF $F, G$ are continuous functions and $F(s)=G(s)$ for all $s>a$, then BOTH $F(s), G(s)$ must behave similarly as $s\rightarrow a^+$ ?

It seems to me the above should be obviously true, but since other OP's strongly disagreed on that post, can someone explain why the above is not necessarily true, if at all it is false ?

Note that though the motivation is the linked post, the present is a genuine question on something that i thought should be obviously true. So this post has really nothing to do with the linked one claiming to prove the RH.

EDIT 1: By ''behave similarly'', i mean if $F(s)$ converges/diverges as $s\rightarrow a^+$, then so must $G(s)$.

For example, let $F(s)=\prod_{p}(1-p^{-s})^{-1}$ denote the Euler product over primes $p$. Then we have $F(s)=\sum_{n=1}^{\infty} n^{-s}$ for ALL $s>1$. Since the sum $1+1/2 + 1/3 + \cdots$ diverges, it follows that $F(s)$ must also diverge as $s\rightarrow 1^+$, and this way one can deduce the infinitude of primes.

EDIT 2: Could the downvotes be explained please ?? Because this post clearly doesn't violate any MSE rule(s). Moderators intervene please.

Answer 1

An interpretation of the question in current form, and something lifted from the comments...

If $F,G$ are continuous functions defined on some common open neighbourhood $U=(a,a+C)\subset \mathbb R$ and $F(s) = G(s)$ for all $s>a$, $s\in U$, [...] if $F(s)$ converges or diverges as $s\to a^+$, then so must $G(s)$.

Yes. Afterall, the definition of the limit as $s\to a^+$ depends only on the values of the function on a right neighbourhood of $a$.

AFTER establishing that the domain of the relevant equation on the real axis includes all $s ≥ Θ$ , which allows one to investigate what happens as $s → ( Θ − \epsilon ) ^+$ .

No. Maybe try substituting actual numbers into your argument. Say that a Big Man in The Sky told you that $\Theta=0$. (For instance, you could be studying the function $\sum_{k=0}^\infty (1-s)^n$ and you're delighted to find out that its equal to $\frac{1}{s}$ for every $2>s>0$.)

Now you want to study a limit as $s\to -\epsilon^+$. Maybe $\epsilon=\frac1{100}$. Maybe smaller, it doesn't matter. I invite you to notice that (and if you got this far, you already agreed) the definition of a limit as $s\to (-\frac{1}{100})^+$, depends on the values of the function on a right neighbourhood of $-\frac{1}{100}$. And there is an entire open interval $(-1/100,0)$ between $-1/100$ and $0$...