Say you are allowed to use Fourier analysis, complex variables, Euler-Maclaurin, etc., but no complex analysis - no holomorphic continuations, no definition of analytic function, and, in particular, no recourse to the concept that every analytic function $f(s)$ vanishing at $s_0$ behaves like $(s-s_0)$ near $s_0$. (You are still allowed to use that fact for a specific function $f$, if you can prove it for that function $f$.)

How would you prove that $lim_{\sigma->1^+} \zeta(\sigma+it) \ne 0$ ($t$ real and fixed)? All the proofs I know (with or without explicit recourse to $\zeta^3(\sigma) |\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)|\geq 0$ or the like) use the fact that, if the limit were 0, then $\zeta(\sigma+it)\sim (\sigma+it-1)$ for $\sigma$ near 1.

(Motivation: of course, I am trying to present a proof of the prime number theorem with plenty of analytic ideas but no complex analysis.)

Say you are allowed to use Fourier analysis, complex variables, Euler-Maclaurin, etc., but no complex analysis - no holomorphic continuations, no definition of analytic function, and, in particular, no recourse to the concept that every analytic function $f(s)$ vanishing at $s_0$ behaves like $(s-s_0)$ near $s_0$. (You are still allowed to use that fact for a specific function $f$, if you can prove it for that function $f$.)

How would you prove that $\lim_{\sigma\to1^+} \zeta(\sigma+it) \ne 0$ ($t$ real and fixed)? All the proofs I know (with or without explicit recourse to $\zeta^3(\sigma) |\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)|\geq 0$ or the like) use the fact that, if the limit were 0, then $\zeta(\sigma+it)\sim (\sigma+it-1)$ for $\sigma$ near 1.

(Motivation: of course, I am trying to present a proof of the prime number theorem with plenty of analytic ideas but no complex analysis.)