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.

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.