Trying to figure out the asymptotic density

Question

Let $a>0$.

I have the following function $f: \mathbb{N} \to \mathbb{R} $ defined in the following way:

\begin{equation}\label{km relation} f(m) = m +\lfloor a m - \tfrac{a }{2} +\tfrac12 \rfloor. \end{equation}

Now, define a `counting function' $A: \mathbb{R} \to \mathbb{R}$ by \begin{equation}\label{} A(x) = \sum_{\substack{a \in f(\mathbb{N}) \\ 1 \le a \le x}} 1. \end{equation}

where $f(\mathbb{N})$ is the image of $\mathbb{N}$ under $f$.

What is $\lim_{x \to \infty} \tfrac{A(x)}{x}$?

I think that it should be $\frac{1}{a+1}$, but since $a$ can be irrational, I can't seem to quite prove it- I get stuck.

Answer 1

Show that $$\lim_{n\to \infty} \frac{f(n)}n=a+1,\qquad \lim_{x\to \infty} \frac{f(A(x))}{x}=1$$

$$\lim_{x\to \infty} \frac{f(A(x))}{x}=\lim_{x\to \infty} \frac{f(A(x))}{A(x)}\frac{A(x)}{x}=\lim_{x\to \infty} (a+1)\frac{A(x)}{x}$$