Limit problem involving bijective function Let $f: \Bbb N^{\star}\to \Bbb N^{\star}$ a bijective functions such that exists $$\lim _{n \to \infty} {\frac {f(n)} {n}}.$$ Find the value of this limit.
I noticed that, if $f(n)=n$, then the limit is $1$. I couldn't make more progress. Can you help me?
 A: Let $f$ be permutation of $\mathbb{N^*}$. Suppose the inequality $f(n) \leq n$ holds for only a finite set of integers $A$. Let $m = \max A$, then $f \left( [m + 1,\infty) \cap \mathbb{N} \right ) \subset [m + 2,\infty) \cap \mathbb{N}$, thus $f^{-1} \big ( [\![ 1,m+1 ]\!] \big ) \subset [\![ 1,m ]\!]$, which is not possible.
Thus $f(n) \over {n}$ $ \leq 1$ holds infinitely many times.
Applying the same reasoning to $f^{-1}$ we get :  $\frac{f^{-1}(n)}{n} \leq 1 $ i.e $f(n) \over {n}$ $ \geq 1$ holds infinitely many times.
Since the limit exists, $\lim \frac{f(n)}{n} = 1$ .
A: Let $L = \lim_{n \to \infty} \frac{f(n)}{n}$. Note $L \geq 0$. Consider the cases:
$L > 1$:
Let $\epsilon > 0$ be such that $L > 1 + \epsilon$. Then $\exists N \in \Bbb N$ such that $\frac{f(n)}{n} > 1 + \epsilon \quad \forall n \geq N$ $\iff f(n) > (1+\epsilon)n$.
Take $M > N$ sufficiently large such that $\epsilon M > 1 \iff (1 + \epsilon)M > M + 1$. Then $\forall n > M, f(n) > M + 1$, so all of $f^{-1}(1), f^{-1}(2), \dots f^{-1}(M + 1)$ must be given by $n_1, n_2, \dots, n_{M+1} \leq M$, but then by the pigeon-hole principle, at least two of the $n_i$ are equal which is a contradiction, as $f$ is a bijection.
$0 \leq L < 1$:
Take $\epsilon > 0$ such that $L < 1 - \epsilon$. Proceed similarly to the $L > 1$ case.  $\exists N \in \Bbb N$ such that $\frac{f(n)}{n} < 1 - \epsilon \quad \forall n \geq N \iff f(n) < (1-\epsilon)n$. Choose $M > N$ sufficiently large such that $\epsilon M > 1 \iff (1-\epsilon)M < M - 1$. Then $\forall n > M, f(n) < M - 1$, but this is clearly a contradiction since $f$ is a bijection.
Hence $L = 1$.
A: Because this limit is finite, exist $M$ and $n_0$ such that $f(n)/n<M$ for all $n>n_0$ ie is bounded ae.
Now, because $f(n)<Mn $ for all $n>n_0$ then $f=O(n)$ and by definition the limit request is 1
