Can the following simple "squeeze theorem" be used to show the Riemann zeta function's singularity is a simple pole of order 1, with the derivation being done entirely in the real domain?
The method is inspired by a question: Why does $\frac{s}{s-1} > \zeta(s) > \frac{1}{s-1}$ imply $\lim_{s \to 1^{+}}(s-1)\zeta(s)=1$?
Step 1
The inequality is derived in the real domain by comparing the sum $\sum1/n^s$ to the integral $\int 1/x^sdx$.
$$\boxed{\frac{1}{s-1}<\zeta(s)<\frac{1}{s-1}+1}$$
The derivation (eg here) is for real $s$.
Step 2
It is then simple algebra to show
$$1<(s-1)\zeta(s)<s$$
which, again, is for real $s$.
As $s\rightarrow 1^+$,
$$1<(s-1)\zeta(s)<1$$
Step 3
We know $\zeta(s)$ has a singularity at $s=1$, so the above shows it is a removable pole of order 1, a simple pole.
Question:
This analysis was done entirely in the real domain. Is the result valid when we consider $s$ to be complex?
If yes, this seems an easy to way to show the Riemann Zeta function's pole is of order 1.
No, that argument is not valid. If $a \in \Bbb R$ and $f$ is holomorphic in $B_r(a) \setminus \{ a \}$ then the existence of $$ \lim_{x \to a, x \in \Bbb R} (x-a)f(x) $$ does not imply that $f$ has simple pole at $a$. A counterexample is $$ f(z) = \frac 1z + e^{-1/z^2} $$ which has an essential singularity at $z=0$.