1. Assume there exists $N$ such that $\phi(n) > n$ for all $n\geq N$. Then $\phi(n) > N$ for all $n\geq N$, leaving not enough values to fill the set $\{1,2,\ldots,N\}$. Therefore for any $N$ there exists some $n\geq N$ such that $\phi(n)\leq n$, and so $\liminf_{n\to \infty} \frac {\phi(n)} {n} \leq 1$. 2. Assume there exists $N$ such that $\phi(n) < n$ for all $n\geq N$. Let $m = \max \{\phi(k) \mid 1\leq k\leq N\}$. Clearly $m \geq N$, but then $\max \{\phi(k) \mid 1\leq k\leq m+1\} = m$, absurd. Therefore for any $N$ there exists some $n\geq N$ such that $\phi(n)\geq n$, and so $\limsup_{n\to \infty} \frac {\phi(n)} {n} \geq 1$. 3. So, if $\lim_{n\to \infty} \frac {\phi(n)} {n}$ does exist, it can only be equal to $1$. A trivial example is for $\phi(n) = n$ for $n\geq 1$, but there are other examples, like $\phi(2n-1) = 2n$, $\phi(2n) = 2n-1$ for $n\geq 1$.