Little o notation confusion

[soggybanana]

I don't understand in this notation of the Prime Number Theorem what is f(x) or g(x) for their little o notation:

"The Prime Number Theorem states the number of primes at most x is Li(x) + o(Li(x)) where Li(x) is integral from 2 to x dt/logt and the notation f(x) = og(x) means lim x-> infinity f(x)/g(x) = 0."

I understand that they are using the offset logarithmic integral notation with Li(x) but I am confused of the little o notation with what is f(x) or g(x). If Li(x) is then f(x) what is what is g(x)? I think I am missing something very simple probably.

 

[HallsofIvy]

f(x) is "o(g(x))" if and only if \(\displaystyle \lim_{x\to\infty} \frac{f(x)}{g(x)}=0\). That is saying that the "number of primes at most x" is Li(x) plus an "error" that goes to 0 as x gets larger.

 

[Dr.Peterson]

I don't understand in this notation of the Prime Number Theorem what is f(x) or g(x) for their little o notation:

"The Prime Number Theorem states the number of primes at most x is Li(x) + o(Li(x)) where Li(x) is integral from 2 to x dt/logt and the notation f(x) = og(x) means lim x-> infinity f(x)/g(x) = 0."

I understand that they are using the offset logarithmic integral notation with Li(x) but I am confused of the little o notation with what is f(x) or g(x). If Li(x) is then f(x) what is what is g(x)? I think I am missing something very simple probably.

We can expand the statement like this, if it helps:

The Prime Number Theorem states the number of primes at most x is Li(x) + E(x) where Li(x) is integral from 2 to x dt/logt and E(x) = o(Li(x)), where the notation f(x) = og(x) means lim x-> infinity f(x)/g(x) = 0.​

So f(x) is the error, E(x), and g(x) is Li(x). The error grows more slowly than Li. (By the way, this doesn't necessarily mean that the error itself goes to zero, just that the ratio does.)