floor function inequality $n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x$ I have the following problem which has two parts:
Given any $x>0$, I define a function $A$ by
\begin{equation}
A(x)=\max \{ n \in \mathbb{N} : n+ \left \lfloor \tfrac{n}{3} \right \rfloor  \le x\}
\end{equation}
I want to compute $A(x)$ and
\begin{equation}
\lim_{x \to \infty} \dfrac{A(x)}{x}. 
\end{equation}
I tried to use the fact that $ x \le \lfloor x \rfloor \le x+1 $ but I got stuck.
Do I have to do a case by case analysis of if $n$ is divisible by $3$?

Since $n\in\mathbb{N}$, it follows by contradiction that
$$n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq x \iff n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq \lfloor x\rfloor$$
Suppose that it is possible to make the above equation an equality. Writing $n=3k+r,0\leq r\leq 2$ we note that
$$\lfloor x\rfloor=4k+r$$
Thus the only cases where such an equality is allowed is when $\lfloor x\rfloor\equiv 0,1,2 \mod 4$.  Explicitly, in this case we can write the solution $n=\lfloor x\rfloor-\lfloor x/4\rfloor$.
Now when $x=3\mod 4$ we see that, since we cannot achieve equality with $\lfloor x\rfloor$ we can move to the closest integer to that, $\lfloor x \rfloor-1.$ We can certainly achieve equality here, with $n=\frac{3(\lfloor x \rfloor-1)+2}{4}=\lfloor x\rfloor-\lfloor x/4\rfloor-1$. Thus
$$A(x)=\begin{Bmatrix}\lfloor x\rfloor-\lfloor x/4\rfloor&x\neq 3\mod 4\\ \lfloor x\rfloor-\lfloor x/4\rfloor-1&x= 3\mod 4 \end{Bmatrix}$$
It should be clear that
$$\lim_{x\to\infty}\frac{A(x)}{x}=\frac{3}{4}$$
EDIT: It turns out a more elegant form exists:
$$A(x)=\lfloor x\rfloor-\left\lfloor \frac{x+1}{4}\right\rfloor$$
One can see this by noticing that the following Iverson bracket, the difference between the two branches above,  is succintly given by
$$[x=3\mod 4]=\left\lfloor \frac{x}{4}+\frac{1}{4}\right\rfloor-\left\lfloor \frac{x}{4}\right\rfloor$$
