I came across the following result in a number theory book. In proving a result of Chebychev that there exist positive constants $x_0$, $c_1$ and $c_2$ such that $$ c_1 \frac{x}{\log x} \le \pi(x) \le c_2\frac{x}{\log x}$$ for all $x>x_0$, the following bounds on $\psi(x)$ were used $$ x\log 2 + O(\log x) \le \psi(x) \le 2x\log 2 + O(\log^2(x))$$ together with the result that $$\frac{\psi(x)}{x} = \frac{\pi(x)\log x}{x} + o(1)$$
The author says that for any $c_1 < \log 2$ and any $c_2 > 2\log 2$ the theorem holds. In an attempt to fill the details, I did the following:
starting from $$ x\log 2 + O(\log x) \le \psi(x) \le 2x\log 2 + O(\log^2(x)),$$ divide through by $x$ to obtain $$\log 2 + o(1) \le \frac{\psi(x)}{x} \le 2\log2 + o(1)$$
Now, replace $\frac{\psi(x)}{x}$ with $\frac{\pi(x)\log x}{x} + o(1)$ so that $$\log 2 + o(1) \le \frac{\pi(x)\log x}{x} + o(1) \le 2\log2 + o(1)$$ Here's a step I'm not sure of and need clarification: can the little $o$ term in the middle be subtracted across the board to get $$\log 2 + o(1) \le \frac{\pi(x)\log x}{x}\le 2\log2 + o(1)$$ $$\frac{x}{\log x}(\log 2 + o(1)) \le \pi(x) \le \frac{x}{\log x}(2\log2 + o(1))$$ and hence for any $c_1 < \log2, c_2 > 2\log 2$ the theorem follows?
Secondly, the author says that since $\frac{2\log 2}{\log 2}=2,$ a corollary of this result is that for every $\epsilon > 0$, there exists a prime number in the interval $[x, (2+\epsilon)x]$ for all $x > x_0(\epsilon)$. I do not see this immediately. How can this be verified?
Thanks to @GregMartin's guidance, here's how to verify the last claim that was made.
Suppose for some $\epsilon > 0$ there is no prime in the interval $[x, (2+\epsilon)x]$ for some $x$. Then, $$\begin{align} 0 = \pi((2+\epsilon)x) - \pi(x) &\ge \frac{c_1(2+\epsilon)x}{\log((2+\epsilon)x)} - \frac{c_2x}{\log x} \\ &= \frac{c_1(2+\epsilon)x}{K + \log x} - \frac{c_2x}{\log x}, \ \ \text{ where } K=\log(2+\epsilon) \\ &= \frac{x}{\log x} \left(\frac{c_1(2+\epsilon)}{1+K/\log x} - c_2\right) \\ &= \frac{x}{\log x} \left(\frac{c_1(2+\epsilon)}{1+o(1)} - c_2\right) \ \ \text{ for } x \text{ large}\\ &= \frac{x}{\log x} \left(c_1(2+\epsilon) +o(1) - c_2\right) \\ &= \frac{x}{\log x}(c_1\epsilon + o(1)) \ \ \text{ since } 2c_1 - c_2 = o(1) \\ \end{align}$$ which is positive as $x \rightarrow \infty$, a contradiction.