Growth of $\pi(2x) - 2\pi(x)$

Question

In Hardy & Wright's Theory of Numbers (p. 494f in 6^th ed.) there's a little discussion following the proof of the prime number theorem.

We have $$ \pi(2x) - \pi(x) = \frac{x}{\log x} + o\left(\frac{x}{\log x}\right) \sim \pi(x). \tag{1} $$ Thus, to a first approximation, the number of primes between $x$ and $2x$ is the same as the number less than $x$. At first sight this is surprising, since we know that primes near $x$ 'thin out' (in some vague sense) as $x$ increases. In fact, $\pi(2x) - 2\pi(x) \to \infty$ as $x \to \infty$ (though we cannot prove this here), but this is not inconsistent with (1), which is equivalent to $$ \pi(2x) - 2\pi(x) = O(\pi(x)). \tag{2} $$

Isn't this just plain wrong? First of all, (1) is not equivalent to (2) but rather to $$ \pi(2x) - 2\pi(x) = o(\pi(x)). \tag{2'}$$ More importantly, how can $\pi(2x) - 2\pi(x)$ go to infinity if $\pi(2x) < 2\pi(x)$ for $x \ge 11$?

Thus the question is, as $x \to \infty$, what is $\pi(2x) - 2\pi(x)$ actually doing?

Answer 1

One can show via the prime number theorem that \[2\pi(x) - \pi(2x) \sim 2 \log 2 \frac{x}{(\log x)^2},\] so that $2\pi(x) \geq \pi(2x)$ for all sufficiently large $x$. See this answer.

Answer 2

It is known that for any $\varepsilon > 0$ there is a point $x_0$ such that $$\frac{x}{\log x - 1 + \varepsilon} < \pi(x) < \frac{x}{\log x - 1 - \varepsilon}, \quad x \ge x_0$$ (e.g. Rosser (1941) Theorem 29A gives $x_0 = 55$ for $\varepsilon = 3$).

From this we deduce the upper bound $$\pi(2x) - 2\pi(x) < \frac{2x(2\varepsilon - \log 2)}{(\log 2x - 1 - \varepsilon)(\log x - 1 + \varepsilon)}$$ which holds eventually for any positive $\varepsilon$. Fixing $\varepsilon = \frac{1}{3} < \frac{1}{2} \log 2$ (because $e^2 < 2.75^2 = 8 - \frac{7}{16}$) we see that $\lim_{x \to \infty} \pi(2x) - 2\pi(x) = -\infty$, not $+\infty$. So there was a typo in the book.