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

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    with $\times$ you mean $\cdot$, right? – 2017-01-14
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    Correct, multiplication – 2017-01-14
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    For $x > 0$: Supposing $a$ is positive, what can you say about $ a\cdot\lfloor\frac{x}{a}\rfloor - x$? – 2017-01-14
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    Now suppose $a$ is negative. – 2017-01-14
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    @Anonymous $\times$ is a standard notation for multiplication of real numbers. This notation puts emphasis on the *operation* rather than on the *result*. The notations $\cdot$ and $xy$ (just concatenation) are also used but they put emphasis on the *result* rather than on the *operation*. Here maybe the latter notations would be better. – 2017-01-14
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    @NeedForHelp Multiplication was indeed the most natural assumption, but making sure did not hurt. – 2017-01-14
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    Further hint: try answering my earlier questions for $x = 3.51623$, for $a = 1, 0.1, 0.01, 0.001$; edit your question to include your results, so that we can see how you're doing. – 2017-01-14
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    My hint would be to write $a\lfloor \frac{x}{a} \rfloor$ as $a(\frac{x}{a}+y)$, and check what happens to $ay$ as $a$ goes to $0$! – 2017-01-14
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    See also: [Does $\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?$](http://math.stackexchange.com/q/1876062) – 2017-01-14

2 Answers 2

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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.

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    Yeah this solution is the very epiphany I just had – 2017-01-14
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    You have tacitly assumed that $a>0$. So, what happens when $a<0$? – 2017-01-14
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    @Dr.MV Thanks for pointing out the oversight. Fixed now. – 2017-01-14
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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.

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    Thank you Dr. John Hughes for your efforts in helping – 2017-01-14
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    You have tacitly assumed that $a>0$. So, what happens when $a<0$? – 2017-01-14
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    True. The case of $a\lt0$ is very similar though, just reverse the inequality signs in the first step – 2017-01-14