# Why prime number theorem tends to one?

I am curious about prime numbers so i just started reading about it. While reading some articles i came across prime number therom (PMT) which states like this

$$\displaystyle \lim_{n \to \infty} \pi \left({n}\right) \frac {\ln \left({n}\right)} n = 1,$$

but when limit tends to infinity $1/n$ should be zero

$$\lim_{n\to\infty}\left(\dfrac1x\right)=0.$$

So PMT tends to infinity answer should be zero and not one!!

Sorry for asking this basic question but my mathematics is bit rusty. Can you tell me why PMT tends to one at infinity? Advance thanks for your help

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I'm not sure what the confusion is. $\pi(n)$ is a number that varies with $n$: for large $n$ it's of the order of $n/(\ln n)$, which is what the prime number theorem is saying. If it helps, just imagine that $\pi(n) = n/(\ln n)$ exactly for large $n$, and convince yourself why there isn't any contradiction -- or better explain what exactly you think the contradiction is. – ShreevatsaR Jun 12 '13 at 5:32
Where does your "So PMT tends to infinity answer should be zero and not one!!" come from? $\lim \frac{1}{n}$ and $\lim \frac{\pi(n) \ln n}{n}$ are two different limits. For example, $\lim \frac{1}{n} = 0$, but $\lim \frac{n}{n}$ is clearly 1, and it doesn't follow from $\lim \frac{1}{n} = 0$ in any way. – user27126 Jun 12 '13 at 5:33

The function $\pi(n)$ can grow faster than $1/n$, or slower, or the same rate, so there is no reason a priori to expect any particular limit. Certainly you agree, for example, that $\lim_{n\to\infty}e^n\frac{\ln(n)}n=\infty$, not $0$, because the $e^n$ term grows so fast. So this is really a statement about the otherwise unknown function $\pi(n)$: it grows at exactly the right rate so that $\lim_{n\to\infty}\pi(n)\frac{\ln(n)}n=1$, which is sometimes written $\pi(n)\sim\frac n{\ln(n)}$ to say that for large $n$, $\pi(n)$ and $\frac n{\ln(n)}$ are at the same order of magnitude. The limit formulation is just a way of making this precise.

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+1. Shouldn't the $1/n$ in the first line be $n$? – ShreevatsaR Jun 12 '13 at 5:40
@ShreevatsaR I meant "the $\pi(n)$ term may or may not dominate the $1/n$ term". Not sure how to say that properly. – Mario Carneiro Jun 12 '13 at 5:41