The use of limit in Titchmarsh's book "The theory of the Riemann zeta-function" in Theorem $3.13$ In Titchmarsh's book "The theory of the Riemann Zeta-function" his Lemma $3.12$ is one of the main tools. Lemma $3.12$ is a version of Perron formula. Lemma $3.12$ starts by observing that, with $c>0$ and with $n<x$,
$$
\frac{1}{2\pi i}\int\limits_{c-iT}^{c+iT}\left(\frac{x}{n}\right)^w \frac{\mathrm dw}{w}=1+
O\left( \frac{(x/n)^c}{T\log (x/n)} \right)
$$
This stems from the fact that $c>0$, and so if one completes the integration contour to the left by encircling the point $w=0$ completely in the complex $w$ plane by a rectangle whose right-most side is the original integration line $(c-iT,c+iT)$, the residue is simply $1$. This is the reason for the requirement $c>0$. The requirement $n<x$ comes from letting the left-most side of the integration rectangle be at $\Re(w)=-\infty$ and so $(x/n)^w\to 0$ if and only if $n<x$ as $\Re(w)\to -\infty$.
And so, in Theorem $3.13$ the starting point is the application of Lemma $3.12$,
$$
\sum\limits_{n<x} \frac{\mu(n)}{n^s}=\frac{1}{2\pi i}\int\limits_{c-iT}^{c+iT}
\frac{x^w \mathrm dw}{w\zeta(s+w)}+
O\left( \frac{x^c}{cT} \right)+
O\left( \frac{\log x}{T} \right)
$$
On the LHS the condition $n<x$ is the condition from Lemma $3.12$, and $c$ is still required to obey $c>0$.
Now Titchmarsh enlarges the contour of integration and encloses the $w=0$ completely, as earlier. Do notice that $c>0$ should still hold, since $c$ is nothing else but $\Re(w)$ on the right-most side the integration contour. No matter how the original line $(c-iT,c+iT)$ is deformed, the deformed closed integration contour must enclose the point $w=0$, and hence $c>0$ on the RHS part of the contour that intersects the real axis in $w$ plane.
With $c>0$ the pole at $w=0$ produces the residue $1/\zeta(s)$ and the result is now
$$\sum\limits_{n<x} \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}+O$$
The $O$ term tends to $0$ in the end, after Titchmarsh adjusts $T$ and $x$ appropriately. 
And so then Titchmarsh requires
$$c=\frac{1}{\log x}$$
Additionally, he then lets $x\to\infty$. 
But in this limit, $c\to 0$. And this makes the original starting equation useless, since the pole of the integrand at $w=0$ is no longer enclosed in the integration countour, but lies on the integration contour instead, if one interprets the limit as $\lim\limits_{x\to\infty}x=\infty$. 
On the other hand, if one interprets the limit $\lim\limits_{x\to\infty}x$ in the way that $x$ grows large but never hits the point at infinity, then $c$ never reaches $0$, and this would do. But aren't there problems with this interpretation of a limit in standard mathematics? 
Maybe Titchmarsh uses the Sokhotski–Plemelj theorem, but then the result misses the summand $\frac{1}{2\zeta(s)}$...
Or maybe I'm missing some detail completely here...
So my question is:
What exactly enables Titchmarsh to take the limit $x\to\infty$?
 A: Let $\epsilon >0$; by the above (standard stuff you seem to agree with), there is a large enough half-integer $x(\epsilon)$ (depending on $s=1+it, \epsilon$), s.t. $|\sum\limits_{n<x} \frac{\mu(n)}{n^s}-\frac{1}{\zeta(s)}| < \epsilon$ for any other half-integer $x \geq x(\epsilon)$ - as usual there is no restriction in using half-integers (i.e. $odd/2$) beacause the Dirichlet sum is constant between integers and the jump goes to zero since $s=1+it$
(choosing in Perron, $c=\frac{1}{\log x}, \log T= (\log x)^{\frac{1}{10}}, \delta = A(\log T)^{-9}=A(\log x)^{-\frac{9}{10}},x$, hence $T$ large enough, $A$ positive absolute constant coming from zero-free regions of $\zeta$)
But then the relation $|\sum\limits_{n<x} \frac{\mu(n)}{n^s}-\frac{1}{\zeta(s)}| < \epsilon$ for $x>x(\epsilon)$ is precisely what is needed to conclude that the limit as $x$ goes to $\infty$ of $\sum\limits_{n<x} \frac{\mu(n)}{n^s}$ is precisely $\frac{1}{\zeta(s)}$, by the usual definition of limit.
So the point is that you apply Perron with large but finite $x$ and then in "shorthand" you let $x$ go to $\infty$ meaning the above $\epsilon - x(\epsilon)$ limit relation, which has nothing to do with making $x$ infinite as a number
A: To see the idea of the proof, let $M(x) = \sum_{n \le x} \mu(n)$ and we'll look instead at  $f(x) = \int_1^x M(y)dy$ to obtain absolutely convergent integrals. 
Then Titchmarsh shows a lower bound $|\zeta(1+it)|\ge B/\log(2+| t|)$ plus an upper bound $|\zeta'(s)|\le B\log(2+| t|)$ to obtain some bound $|\zeta(s)|>1/(A\log(2+|t|)) ,|\frac1{\zeta(s)}|<\log(2+|t|)$ for $s=\sigma+it, \sigma\ge 1-\frac{1}{A\log^2(2+ |t|)}$. 
And this is what we need to conclude $$f(x) = \frac{1}{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{1}{\zeta(s)} \frac{x^{s+1}}{s(s+1)}ds= \frac{1}{2i\pi} \int_{\Re(s) = 1-\frac{1}{A\log^2 (2+|\Im(s)|)}} \frac{1}{\zeta(s)} \frac{x^{s+1}}{s(s+1)}ds \\= O(\int_{-\infty}^\infty \frac{\log (2+|t|)}{1+t^2 }x^{2-\frac{1}{A \log^2(1+ |t|)}}dt)\\ = 
O(\int_0^T \frac{x^{2-\frac{1}{A \log^2(1+ |T|)}}\log (2+|t|)}{1+t^2 }dt)+O(\int_T^\infty \frac{x^{2}\log T}{1+t^2 }dt)\\
=O(x^{2-\frac{1}{A \log^2(1+ |T|)}}) + O(\frac{x^2 \log T}{T})\\=O(x^{2-\frac{1}{A \log^2(1+ e^{\log^{1/4} x})}}) +O(\frac{x^2\log^{1/4} x}{e^{\log^{1/4} x}})=O(\frac{x^2}{e^{\log^{1/8} x}})=o(x)$$
If $M(x)> cx$ infinitely often, as $M(x+y)\ge M(x)-y$ then $|f(x+cx/2)-f(x) | \ge \sum_{n=0}^{cx/2} (cx-n) \ge x^2 c^2/8$ infinitely often, contradicting that $f(x)=o(x^2)$. Thus we proved $M(x)=o(x)$, the PNT.
