Limit of the Euler totient function

[henpen]

My question is relatively breif: is it true that

$$\displaystyle \lim_{n \rightarrow \infty}(\varphi(n))=\lim_{n \rightarrow \infty}(n) \cdot \prod_{i=1}^{\infty}(1-\frac{1}{p_i})$$
Where $p$ is prime? Pehaps $\varphi(n)$ is too discontinuous to take the limit of, but it would seem that as it increases to infinity the function should tend to infinity, with fewer anomalies.

If this were true,

$$\displaystyle \zeta(1)=\frac{1}{ \prod_{i=1}^{\infty}(1-\frac{1}{p_i})}=\frac{1}{\lim_{n \rightarrow \infty}(\frac{\varphi(n)}{n})}=\lim_{n \rightarrow \infty}(\frac{n}{\varphi(n)})$$

 

[Mandelbroth]

My question is relatively brief: is it true that

$$\displaystyle \lim_{n \rightarrow \infty}(\varphi(n))=\lim_{n \rightarrow \infty}(n) \cdot \prod_{i=1}^{\infty}(1-\frac{1}{p_i})$$
Where $p$ is prime? Pehaps $\varphi(n)$ is too discontinuous to take the limit of, but it would seem that as it increases to infinity the function should tend to infinity, with fewer anomalies.

If this were true,

$$\displaystyle \zeta(1)=\frac{1}{ \prod_{i=1}^{\infty}(1-\frac{1}{p_i})}=\frac{1}{\lim_{n \rightarrow \infty}(\frac{\varphi(n)}{n})}=\lim_{n \rightarrow \infty}(\frac{n}{\varphi(n)})$$

Let $\mathfrak{P}_n$ be the set of all distinct prime divisors of a number n.
Consider that $\displaystyle \varphi(n) = n \prod_{i=1}^{\sharp\mathfrak{P}_n}\left[1-\frac{1}{p_i}\right]$, where $\sharp\mathfrak{P}_n$ is the cardinality of $\mathfrak{P}_n$ and $p_i$ is the i^th element of $\mathfrak{P}_n$. As n increases, its number of prime factors tends to increase, but this trend is in no way strictly true for individual numbers. An example of a relatively large number that does not have a large number of prime factors is 87178291199, which has only one prime factor. :tongue:

Thus, using basic properties of limits, your formula should be correct.

 

[henpen]

The problem I has was that if particular numbers are 'discontinuous' from the general trend, you can't take the limit, even if the general trend tends to infinity.

 

[Mandelbroth]

The problem I has was that if particular numbers are 'discontinuous' from the general trend, you can't take the limit, even if the general trend tends to infinity.

Consider a function $f: \mathbb{R}\rightarrow\mathbb{R}\cup\left\{Tootsiepop\right\}$, where, for $x\in\mathbb{R}$, $f(x)=\left\{\begin{array} , x , x\neq2 \\ Tootsiepop , x=2 \end{array}\right.$

As x approaches 2, f(x) approaches 2. However, f(2)=Tootsiepop.

This is not a continuous function, but the limit as x approaches 2 is defined. So, I don't understand what you mean...

 

[henpen]

The example cleared up a lot, thanks. I've little formal experience with limits.