$\liminf_{n\to\infty} \frac{\varphi(n)}{n} = 1$, not $0$ Let $\varphi(n) = \sharp\{1\leq x \leq n : (x,n) = 1\}$.  Then $\liminf_{n\to\infty} \frac{\varphi(n)}{n} = 1$.   My attempt: $\inf_{k\geq n}\frac{\varphi(k)}{k} \leq 1$ since for $n = 2$, this infimum equals $1$, we have that $\sup_{n\geq 1} \inf_{k\geq n} \frac{\varphi(k)}{k} = 1$.
Okay, that proof didn't work because $\varphi(2)/2 = 0.5$, but letting $n = 1$, we have $\varphi(1)/1 = 1$.  QED
Is that correct?  Then my book is wrong:
Probabilistic Number Theory by Dr.J¨orn Steuding (page 4)
 A: About lim inf, if we can provide a sequence of numbers $n$ such that $\phi(n)/n$ goes to zero, then your liminf is, in fact, zero.
The sequence we use is the primorials, the products of the consecutive primes up to some bound. So, the sequence is $$ 2,6,30, 210,\ldots $$
If $$ N = \prod_{p \leq x} p $$ for some postive real $x,$ then 
$$ \frac{\phi(N)}{N} = \prod_{p \leq x} \left( 1 - \frac{1}{p} \right).  $$
As $x \rightarrow \infty,$ this product goes to $0.$ Taking logarithms, this follows from the fact that the harmonic sum of the primes diverges. Indeed, for $x > 1,$ from Rosser and Schoenfeld (1962), (3.26) on page 70, we have
$$ \frac{\phi(N)}{N} = \prod_{p \leq x} \left( 1 - \frac{1}{p} \right) < \frac{e^{-\gamma}}{\log x} \left(1 + \frac{1}{2 \log^2 x} \right).  $$
Here $\gamma = 0.5772156649...$ is the Euler-Mascheroni constant.
A: Let $(a_{n})$ be a sequence in $\mathbb{R}$ and let $l \in \mathbb{R}$. Then it can be shown that $l = \liminf_{n \to \infty}a_{n}$ if and only if for every $\varepsilon > 0$, i) there is some $N \geq 1$ such that $\inf_{n \geq N}a_{n} > l-\varepsilon$ and ii) for every $N \geq 1$ there is some $n \geq N$ such that $a_{n} < l + \varepsilon$.
Let $a_{n} := \varphi(n)/n$ for all $n \geq 1$. We claim that $\liminf a_{n} = 0$. Since $a_{n} > 0$ for all $n \geq 1$, Property i) holds for every $\varepsilon > 0$. But because of the prime form of $a_{n}$, for every $\varepsilon > 0$ we have $a_{n} < \varepsilon$ for large $n$, and Property ii) follows.
Thus $\liminf a_{n} = 0$.
