Limit $\lim\limits_{a \to 0}( a\lfloor\frac{x}{a}\rfloor)$

Question

Obviously the limit either does not exist or converges to $x$.
I'm partial towards the latter, and have an incomplete argument involving Fourier Series which corroborates my inclination. Feel free to disprove me/affirm my hunch. $$\lim\limits_{a \to 0}( a\lfloor\frac{x}{a}\rfloor)$$

Answer 1

We know that $\lfloor x/a \rfloor \le x/a < \lfloor x/a \rfloor +1$, so $$ a > 0 \Rightarrow a\lfloor x/a \rfloor \le x < a\lfloor x/a \rfloor +a \\ a < 0 \Rightarrow a\lfloor x/a \rfloor +a < x \le a\lfloor x/a \rfloor < $$ which means that $$ a > 0 \Rightarrow 0 \le x - a\lfloor x/a \rfloor < a \\ a < 0 \Rightarrow a < x - a\lfloor x/a \rfloor \le 0, $$ and so $$ | x - a\lfloor x/a \rfloor | \le |a|. $$ Thus $a \lfloor x/a \rfloor \to x$ as $a \to 0$ by the squeeze lemma.

Answer 2

Overthought it:
For $a\gt0$:
$\frac{x}{a}-1 \lt \lfloor\frac{x}{a}\rfloor \lt \frac{x}{a}+1$
$a\frac{x}{a}-a \lt a\lfloor\frac{x}{a}\rfloor \lt a\frac{x}{a}+a$
The convergence follows from the squeeze theorem.
$a\lt0$ is similar.