Limit involving prime and composite numbers

Let $p(x) = \frac{x}{\ln x}$, being the approximate prime counting function. That means there are approximately $\frac{x}{\ln x}$ primes less than or equal to x, and $x-\frac{x}{\ln x}$ composites less or equal to than x.

First, let's derive $p_n$. Since there are about $\frac{x}{\ln x}$ primes less than or equal to x, there are x primes less than or equal to $x\ln(x)$. So $p_n \sim n\ln n$. Now for $c_n$, there are $x(1-\frac{1}{\ln x})$ composites less than or equal to x, so there are x composites less than or equal to $\frac{x\ln x}{(\ln x) - 1}$. Therefore, $c_n \sim \frac{n\ln n}{(\ln n) - 1}$

So, your limit now becomes: $\frac{n\ln n}{n(\ln (n) - 1)} - \frac{1}{(\ln n) - 1} - \frac{n\ln n}{n^2(\ln (n) - 1)}$, and as n approaches infinity, the last two terms drop to zero, and a simple coefficient test shows the first term is equal to one. And, $$\lim_{n\to\infty} \frac {c_n}n-\frac{c_n}{p_n}-\frac {c_n}{n^2} = 1$$