Asymptotic limit of truncated Legendre sieve Consider the truncated sum
$$
S(x):=\sum_{\substack{{d\mid P(\sqrt{x})}\\{d\leq x}}}\mu(d)/d,
$$
where $P(z)$ is the product of all primes less than or equal to $z$, and $\mu(d)$ is the Möbius function. 
My question is simply:

Does $S(x)$ have a provable asymptotic limit?

The question is motivated by the following plot:

It is well known from the prime number theorem and Merten's product theorem that 
$$
\frac{x}{\log x \cdot \pi(x)} \sim 1 \quad \textrm{and} \quad
\log x \cdot \sum_{\substack{{d\mid P(\sqrt{x})}}}\mu(d)/d \sim 2 \textrm{e}^{-\gamma}. 
$$
And in seeing the plot, I got curious of
$$
\log x \cdot S(x) \sim ?
$$
It is not possible to state from the numerical example what the asymptotic limit of $\log x \cdot S(x)$ will be. But note that so is the case also of $x/(\log x \cdot \pi(x))$, which provable tends to 1. What I'm curious about therefore, is whether $\log x \cdot S(x)$ will go all the way to 1 or stagnate before that. 
I did check out chapter 4 of Opera de Cribro by Friedlander and Iwaniec. There it appears that at least in terms of the elementary Legendre sieve, it is not possible to prove the asymptotic limit of $S(x)$. It might be that an affirmative answer lies other places in that book, with more powerful sieving methods. I would be glad if anyone could point me to the right place in that case, as it is a rather chunky piece of literature to just browse through.
ADDED 1: Numerically, it appears that the constant $\sum_{k=1}^{\infty} M(k) \log \Big(1+\frac 1k\Big)$ that was stated below in Lucia's answer should be 1, as according to Terry Tao's comment. This is evidenced in the following plot:

ADDED 2: One could also anticipate the asymptotic limit 
$$
S(x)\sim 
\frac{1}{\log x}
$$
from the following sloppy heuristic: $S(x)$ can in some sense be seen as the mean value of prime densities below $x$. If we take those densities to be $1/\log x$ around $x$, the average density is simply $\textrm{li}(x)/x$. We know of course that this expression satisfies 
$$
\frac{\textrm{li}(x)}{x} \sim \frac{1}{\log x}.
$$
We might therefore anticipate $S(x)\sim 1/ \log x$. I have added a new version below of the first plot, that also includes the term $\log x \cdot \textrm{li}(x)/x$, revealing that this lies above $\log x \cdot S(x)$. Naturally, we would not expect the two to coincide, as $S(x)$ must interpreted as the mean of a random model where the actual prime distribution is just one specific outcome. 

 A: One can prove that $(\log x)S(x)$ does tend to a limit, but I don't think the limit is necessarily a nice constant. [Edit The limit is in fact $1$ as noted by Terry Tao and user45947. ] Put $M(z) = \sum_{n\le z} \mu(n)/n$.  Note that by the prime number theorem $M(z) \ll (\log z)^{-A}$ for any $A>0$ and $z$ large enough.  I'll prove that 
$$ 
S(x) \sim \frac{1}{\log x} \sum_{k=1}^{\infty} M(k) \log \Big(1+\frac 1k\Big),
$$ 
and the bound on $M(k)$ guarantees that the sum above is convergent.  This constant can of course be computed (and should be checked with your numerics) but I don't see why it should be anything nice. [ Edit In fact one can see that 
$$ 
\sum_{k=1}^{K} M(k) \log \Big( 1+ \frac 1k \Big) = \sum_{n\le K} \frac{\mu(n)}{n}\sum_{k=n}^{K}\log (1+1/k) = \sum_{n\le K} \frac{\mu(n)}{n} \log \frac{K+1}{n}.
$$ 
Since $\sum_{n=1}^{K} \mu(n)/n = O((\log K)^{-A})$ and $-\sum_{n=1}^{\infty} \mu(n)(\log n)/n=1$, it follows that the constant is in fact $1$. ]
To prove the asymptotic formula, note that $S(x)$ counts all $n\le x$ except for those $n$ having a prime factor $p$ larger than $\sqrt{x}$.  Thus 
$$ 
S(x) = M(x) - \sum_{p>\sqrt{x}} \sum_{\substack{n\le x \\ p|n }} \frac{\mu(n)}{n}.  
$$
The term $M(x) = O(1/(\log x)^2)$ and so we focus just on the second term above.  Writing $n=mp$ this is 
$$
\sum_{p>\sqrt{x}} \frac{1}{p } \sum_{m\le x/p} \frac{\mu(m)}{m}.
$$ 
Now let $1\le k\le \sqrt{x}$ and group the primes $p$ above according to the ranges $x/(k+1) < p \le x/k$.  Thus the sum above equals
$$
\sum_{1\le k\le \sqrt{x}} M(k) \sum_{x/(k+1) <p\le x/k} \frac{1}{p}.
$$ 
Now use the asymptotics for the sum of the reciprocals of primes to see that the inner sum over $p$ above is 
$$ 
\sim \log \frac{\log (x/k)}{\log (x/(k+1))} \sim \frac{\log (1+1/k)}{\log x}. 
$$
From this the desired asymptotic follows.  
The argument above is a quick sketch, and some details would need to be filled in -- but nothing too hard.  The proof follows ideas of Dress, Iwaniec and Tenenbaum and see that paper for further details in a related calculation. 
