# Limit with n tending to infinity

let,s suppose that we have the limit with n tending to infinity:
$$\frac{f(n)}{g(n)}=1$$ then i suppose that for n tending to infinity we should get:
$$f(n)\rightarrow{g(n)}$$ or what is the same the function f(n) diverges as g(n) as an special case:
$$\pi(n)\rightarrow{n/ln(n)}$$ where Pi is the prime number counting function in number theory

matt grime
Homework Helper
eljose said:
let,s suppose that we have the limit with n tending to infinity:
$$\frac{f(n)}{g(n)}=1$$ then i suppose that for n tending to infinity we should get:
$$f(n)\rightarrow{g(n)}$$ or what is the same the function f(n) diverges as g(n) as an special case:
$$\pi(n)\rightarrow{n/ln(n)}$$ where Pi is the prime number counting function in number theory

What is your question? What you;'ve written doesn't make sense.

Are you attempting to ask: if f and g are asymptotic then does f-g tend to zero? Of course not: we can make two functions that are asymptotic diverge absolutely as fast as you want.

Last edited:
benorin
Homework Helper
Gold Member
As I recall, that $\frac{f(n)}{g(n)}=1\mbox{ as }n\rightarrow\infty$ speaks of the asymptotic relationship between f and g, namely that $f(n) \sim g(n)$ (ref. Asymptotic Notation).

My main question is related with proving (if true) the equality

$$\frac{\int_0^{n}dx(x^p)}{1^p+2^p+3^p+............+n^p}\rightarrow{1}$$

as n tends to infinity $$n\rightarrow{\infty}$$

matt grime