Similar uses of $3+4\cos \phi +2\cos 2\phi $ in the prime number theorem This question is about a step in the usual proof of the prime number theorem. Use is made of the relation 
$3+4\cos \phi +\cos 2\phi =2(1+\cos \phi )^{2}\geq 0,$
a good choice because it gives a convenient number of factors in the expression
$$H(\sigma) = [\zeta(\sigma)(\sigma-1)]^3\left(\frac{|\zeta(\sigma+it)|}{\sigma-1}\right)^4|\zeta(\sigma+2it)|(\sigma-1).$$
So I have seen two arguments regarding this step in the proof and they may both be right. The first is from Jameson's The Prime Number Theorem at p. 107. The idea is to show there are no zeros of $\zeta(s)$ on $\sigma=1.$
He says that showing $H(\sigma)\to 0$ contradicts his previous lemma, that 
$$J(\sigma)=\zeta(\sigma)^3|\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)|\geq1. $$ We know that the middle factor of $H$ as $\sigma\to 1+$ is a derivative and is some finite number. We know that  $\zeta(\sigma)(\sigma-1)\to 1$ as $\sigma\to 1. $ Now Jameson says that $\zeta(\sigma +2it)\to \zeta(1+2it),$ also some finite number. So the "extra" factor $(\sigma-1)$ makes $H\to 0$ contradicting the last inequality above.
The second explanation I saw was that if $H\to 0$ as $\sigma \to 1+$ then the factors of $\zeta(1+2it)(\sigma-1)$ become arbitrarily large, small (respectively), contradicting what we know about $\zeta(1+2it).$ This seems to remove the need to show $J(\sigma)\geq 1.$
(The argument in the linked Wiki article seems to say that $\zeta(x)$ has a simple pole at $\sigma=1$ and therefore $J$ tends to zero, for a contradiction. I fail to follow this unless it's shorthand for one of the above arguments.)

Are both arguments above essentially correct? Is there any basis for preferring one over the other, if I have expressed them correctly?

 A: Both arguments are variants of one and the same argument, just differently phrased.
We know unconditionally that


*

*$J(\sigma) \geqslant 1$ for $\sigma > 1$ and arbitrary $t\in \mathbb{R}\setminus \{0\}$ (by the $3,4,1$-trick),

*$\zeta(\sigma)(\sigma-1) \to 1$ as $\sigma \to 1$,

*$\zeta(\sigma + 2it) \to \zeta(1 + 2it)$ as $\sigma \to 1$, and

*$\sigma - 1 \to 0$ as $\sigma \to 1$.


Hence it follows that
$$\lim_{\sigma \to 1^+} \underbrace{[\zeta(\sigma)(\sigma-1)]^3 \lvert \zeta(\sigma + 2it)\rvert (\sigma - 1)}_{A(\sigma)} = 0.\tag{$\ast$}$$
We can conclude the argument in various ways. Without assuming that $\zeta(1 + it) = 0$, from $H(\sigma) = J(\sigma) \geqslant 1$ and $(\ast)$ it follows that
$$\biggl(\frac{\lvert \zeta(\sigma + it)\rvert}{\sigma - 1}\biggr)^4 = \frac{H(\sigma)}{A(\sigma)}$$
is unbounded, and hence $\zeta(1 + it) \neq 0$, for if $1+it$ were a zero of $\zeta$, then $\frac{\zeta(\sigma + it}{\sigma - 1}$ would remain bounded (it would actually converge to $\zeta'(1+it)$).
Under the assumption that $\zeta(1 + it) = 0$, by the previous observation, it follows that $H(\sigma) \to 0$, contradicting $H(\sigma) = J(\sigma) \geqslant 1$ for $\sigma > 1$. This is the first argument from your question.
Or, without using $(\ast)$, assuming that $\zeta(1+it) = 0$ implies that
$$B(\sigma) = [\zeta(\sigma)(\sigma-1)]^3\biggl(\frac{\lvert\zeta(\sigma+it)\rvert}{\sigma-1}\biggr)^4$$
remains bounded for $\sigma \to 1$, and then $H(\sigma) = J(\sigma) \geqslant 1$ for $\sigma > 1$ implies that $\zeta(\sigma + 2it)$ is unbounded, which contradicts the third bullet point above. This is the second argument from your question.
The second bullet point above follows from the fact that $\zeta$ has a simple pole (with residue $1$) at $1$, and it is essential to the argument, in whichever variant one uses it. 
