For every $\epsilon >0$, show that each of the inequalities $$\prod\limits_{p \leq x} p> e^{(1+\epsilon )x} \text{ and } \prod\limits_{p \leq x} p < e^{(1-\epsilon) x}$$ is false for all sufficiently large $x$.
($p$ is prime and $x \in \mathbb{R}$). This is in Leveque's Fundamentals of Number Theory. Is there a way to show this result using $\pi (x) = \frac{x}{\log x} + O\left(\frac{x}{ \log^2 x}\right)$? If not, how can we proceed/conclude otherwise?
From this post I answered some time ago (mainly for the links and references) we have $$\prod\limits_{p\leq x}p=\left \lfloor x \right \rfloor \#=\prod\limits_{k=1}^{\pi(x)}p_k=e^{\sum\limits_{k=1}^{\pi(x)}\ln{p_k}}=e^{\vartheta (x)} \tag{1}$$ where $\vartheta (x)$ is Chebyshev function with the property that $$\lim\limits_{x\rightarrow\infty}\frac{\vartheta (x)}{x}=1$$ which means, $\forall \varepsilon>0$, $\exists x(\varepsilon)>0$ s.t. $\forall x > x(\varepsilon)$ $$(1-\varepsilon)x<\vartheta (x)<(1+\varepsilon)x \tag{2}$$ or, given $e^x$ is ascending function, from $(1)$
$$e^{(1-\varepsilon)x}<\prod\limits_{p\leq x}p< e^{(1+\varepsilon)x}$$ which will help you to prove the statement by contradiction.