I am examining a certain set of integers, $M$. Let's say, the amount of integers less or equal than $x$ is $S(x)$. Now my studies show that the density of primes in this set $M$ is the double density of primes in the naturals. In formula $$ \operatorname{d}_M(x) \sim 2\operatorname{d}(x) \text{ for } x\to\infty. $$ Let's say $\pi(x)$ is the prime counting function and $\pi_M(x)$ is the amount of primes in $M$ which are $\leq x$. Then we have $$ \operatorname{d}_M(x) = \frac{\pi_M(x)} {\operatorname{S}(x)} \;\; \text{ and } \;\; \operatorname{d}(x) = \frac{\pi(x)}{x}. $$ Now I wonder whether I may do the following: $$ \operatorname{d}_M(x) \sim 2\operatorname{d}(x) \Longleftrightarrow \frac{\pi_M(x)} {\operatorname{S}(x)} \sim 2\frac{\pi(x)}{x} \Longleftrightarrow |\pi_M(x)| \stackrel{?}{\sim} 2\cdot\frac{\pi(x)}{x} \cdot \operatorname{S}(x). $$
Because if I use the definition of $\sim$ I get $$ \lim\limits_{x\to\infty} \frac{\frac{\pi_M(x)} {\operatorname{S}(x)}}{2\cdot\frac{\pi(x)}{x}} = 1, $$ but the term in question would be $$ \lim\limits_{x\to\infty} \pi_M(x) = \lim\limits_{x\to\infty}2\cdot\frac{\pi(x)}{x}\cdot \operatorname{S}(x). $$
I don't know whether this is allowed. Nevertheless I know from the prime theorem that we can estimate $\pi(x)$ by calculating $\frac{\pi(x)}{x}$ which is actually what I want to do here for $\pi_M(x)$