The second Chebyshev function is defined as $$\psi(x):= \sum_{p^k\leq x} \log p.$$ Now let $\psi(x) : R \to R$. Then $\psi(x) \sim x$ if and only if for every $\epsilon > 0$, it is said/can be proved (see lemma 8 on page 9) - $$(1 − \epsilon )x < \psi(x)< (1 + \epsilon )x$$ for all sufficiently large $x$.
Is $\epsilon >1$? How can I prove $$(1 − \epsilon )x < \psi(x) (1 + \epsilon )x ?$$
It would be helpful if you provide a detailed proof or source of a detailed proof.
This follows from the definition of $\sim$. Note that $\psi(x)\sim x$ means $\displaystyle\lim_{x\to\infty}\frac{\psi(x)}{x}=1$. Which means for sufficiently large $x$ you have $$\left|\frac{\psi(x)}{x}-1\right|<\epsilon \implies 1-\epsilon <\frac{\psi(x)}{x}<1+\epsilon$$