I'm going through Stein & Shakarchi's Fourier Analysis, and in Chapter 8 on Dirichlet's theorem, they prove the Euler product formula for the Riemann zeta function (Theorem 1.10, pg. 250) by first observing that
\begin{align} \sum_{n=1}^N \frac{1}{n^s} &\leq \prod_{p \leq N} \frac{1}{1-p^{-s}} \\ &\leq \prod_p \frac{1}{1-p^{-s}}. \end{align}
(arguing via the fundamental theorem of arithmetic), so letting $N$ tend to infinity yields $\displaystyle \sum_{n=1}^\infty \frac{1}{n^s} \leq \prod_p \frac{1}{1-p^{-s}}$. To prove the reverse inequality they note (again invoking FTA) that
$$ \prod_p \left( 1+\frac{1}{p^s}+\cdots+\frac{1}{p^{Ms}} \right) \leq \sum_{n=1}^\infty \frac{1}{n^s}. $$
They again let $M$ tend to infinity, then $N$. This concludes the proof.
They do something similar for $L(s,\chi)$ for $\chi$ a completely multiplicative character (pg. 260).
Is an analytic argument necessary? I'm familiar with the proof that successively multiplies factors of $(1-p^{-s})$ to $\sum \frac{1}{n^s}$ to get $\zeta(s) \cdot \prod_p (1-p^{-s}) = 1$, which does not (apparently) require taking limits. If not, is there a reason they do this? To avoid using the infinitude of primes in their proof?