# Existence of a limit, prime number theorem

Why does one have to prove that the limit below exists, in order to prove the prime number theorem? $$\lim_{x \rightarrow \infty} \frac{\psi(x)}{x}$$ Doesn't the fact that the Chebyshev function is monotonic imply that it does exist? If it doesn't then how does one prove that the limit exists? How would I even try to figure out what the limit superior and inferior are?

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Explain your notation. What is $\psi (x)$ here? Is it Chebyshev function? – glebovg Nov 13 '12 at 0:10
${e^x}$ is monotone but ${\lim _{x \to \infty }}{e^x}/x = \infty$. – glebovg Nov 13 '12 at 0:29
Chebyshev function is not continuous, so lim sup and lim inf are not defined here. – glebovg Nov 13 '12 at 0:38
@glebovg: they may be $+\infty$ or $-\infty$, but $\limsup$ and $\liminf$ always exist. – robjohn Nov 13 '12 at 0:40

Next, I found a simple formula, $$f(x) = \left( 3 + \cos (\log x) \right) \, x \; \; \; \mbox{for} \; \; x \geq 1$$ which is monotonic (take the derivative) but for which $f(x) / x$ has no limit as $x \rightarrow \; +\infty.$
Finally, if you are serious about this, what you want is Hardy and Wright, chapter 22. Theorem 414 on page 341 (of the paperback fifth edition) just says that your ratio has an upper bound and a lower bound for $x \geq 2.$ I think Chebyshev proved that much, but they do not say and they say so in the Chapter Notes on page 373. The Prime Number Theorem itself is Theorem 434 on page 362, the proof is completed by the middle of page 367.