Prime number theorem and how many primes are close to $x$ for sufficiently large $n$ The prime number theorem states:
$$
\lim_{x-> \infty}{\frac{\pi(x)}{\frac{x}{ln(x)}}} = 1
$$
I was trying to get a better understanding on the intuition on that statement and more importantly, I was reading in a summary slide on a crypto course online that " about $(1/\ln x)$ numbers near $x$ are prime." I was wondering what "near" meant and what was the intuition to support that claim (based on the above limit). Basically, I think that understanding that limit better conceptually might be the key to understanding that claim better (proofs that answer my question are welcome, but I feel I might not have the knowledge on number theory to understand them well, so I would prefer good intuitions).
 A: One way to express the prime number theorem intuitively is to say that the probability of picking a prime number in the interval $[1,x]$ is approximately $\frac{1}{\ln x}$ (and converges to that when $x\to\infty$). But that is not the same as "close".
Perhaps this version of Bertrand's postulate works for you:
$$
\pi(2x)-\pi(x) \sim \frac{x}{\ln x}
$$
A: Basically what I understand now is that, what the limit says is that for sufficiently large x, $\frac{x}{ln(x)}$ approximates $\pi(x)$ (or they become of the same "order"). Thus, it states that for sufficiently large x, then the number of prime numbers less than x is approximately $\frac{x}{ln(x)}$. The total numbers between 1 to x are x. Thus, the probability of selecting a prime is $\frac{x}{xln(x)} = \frac{1}{ln(x)}$.
I provided a solution that addressed some of the issue I had, but a better solution (even if its slightly similar) that provides more insight/understanding of why the limit implies  what the probability is will get likes from me and I will accept the answer.
