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So I am currently working in the Apostol Book "Introduction to Analytic Number Theory", and am attempting to prove one of the major results regarding the prime number theorem. Specifically I am trying to show that given: $$\lim_{x \to \infty} \frac{\pi(x)\log x}{x} = 1$$ show that: $$\lim_{x\to \infty} \frac{\vartheta (x)}{x} = 1$$ What I have done so far is this: We know for $x > 2$ that: $$\vartheta (x) = \pi (x) \log x + \int_{2}^{x}\frac{\pi(t)}{ t}dt$$ This a result derived previously in the book. So we must show: $$\lim_{x \to \infty} \frac{\pi(x)\log x +\int_{2}^{x}\frac{\pi(t)}{t}dt}{x} = 1$$ Clearly, it suffices to show that: $$ \lim_{x \to \infty} \frac{1}{x}\int_{2}^{x}\frac{\pi(t)}{ t}dt = 0 $$ Apostol then discusses a bound on the integrand. Specifically, states that: $$\lim_{x \to \infty} \frac{\pi(x)\log x}{x} = 1 \implies \frac{\pi(t)}{t} = \mathcal{O} \left(\frac{1}{\log t} \right) $$ for $t>2$. At first, I was unsure as to where this result comes from, but then I realized one could get it by dividing the $\log x$ from both sides of the limit definition. However, it still does not make sense as to what the result means/wether it is rigorous or not. I guess the question I am asking is, what is the connection between big $\mathcal{O}$ notation and asymptotic notation (limit of a quotient)?

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closed as off-topic by Did, Glorfindel, Leucippus, Shailesh, Daniel W. Farlow Jul 27 '17 at 0:27

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  • $\begingroup$ Did you simply look for a definition of Big-O asymptotics? It should allay all your doubts... $\endgroup$ – Did Jul 26 '17 at 17:27
  • $\begingroup$ He does not need something to raise doubts. He needs something to allay doubts. $\endgroup$ – GEdgar Jul 26 '17 at 17:28
  • $\begingroup$ @GEdgar Indeed. Thanks. $\endgroup$ – Did Jul 26 '17 at 17:30
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    $\begingroup$ @Did Looking back at the definition I feel quite stupid for asking this. It becomes fairly obvious. Thanks! $\endgroup$ – rubikscube09 Jul 26 '17 at 17:43
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You have that $$ \lim_{x\to\infty}\frac{\pi(x)\log x}{x} = 1,\tag{1} $$ and the claim is that $$ \frac{\pi(t)}{t} = \mathcal O\left(\frac{1}{\log t}\right)\qquad\text{as $t\to\infty$,} $$ which is to say that there exists $B>0$ and $M>0$ such that $\pi(t)/t \le B/\log t$ for all $t>M$. Clearly the claim is true if and only if $$ \frac{\pi(t)\log t}{t} \le B \tag{2} $$ for all $t$ sufficiently large. Since you are given $(1)$, you have $(2)$ since eventually the quotient $\frac{\pi(t)\log t}{t}$ is $\le 2$, say.

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This is quite rigorous:

All you need is that the stated limit implies that $\pi(x)\log x/x<M$ for $x$ large enough. ($M$ can be any number greater than $1$.) From this you get $\pi(x)/x<M/\log x$ for $x$ large enough, and that is precisely the meaning of the big $\mathcal{O}$ notation.

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