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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?

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    $\begingroup$ Take logs of both sides. PNT is equivalent to $\sum_{p \le x} \log p \sim x$ $\endgroup$ – mathworker21 Feb 25 '18 at 1:00
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    $\begingroup$ Let $p_n$ be the $n$th prime. The PNT implies $p_n\sim n\log p_n.$ With a little elementary work, this implies $p_n\sim n\log n.$ From that we readily obtain $\sum_{p\leq x}\log p\sim x.$ $\endgroup$ – DanielWainfleet Feb 26 '18 at 1:43
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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.

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