Evaluating the limit $ \lim_{x \to 0} \frac{x}{p} \lfloor \frac{q}{x} \rfloor$ 
$$\lim_{x \to 0} \frac{x}{p} \left\lfloor \frac{q}{x} \right\rfloor =\;\;\; ?$$

My attempt:
$\frac{q}{x}-1 < \lfloor \frac{q}{x} \rfloor \le \frac {q}{x} $    . Then I consider four cases. In case 1 I took x and p both positive and then I multiplied above by $\frac{x}{p} $ and took limit and applied squeeze theorem. In other cases I did the same. My answer is $ \frac{q}{p} $.  
Is the solution correct?
 A: Yes, you're right. Let $p, q, x\in \mathbb R$ and $p\neq 0$, then notice, that $$\frac{q}{x}-1 < \left \lfloor\frac{q}{x} \right \rfloor \leq \frac{q}{x}$$
Now we may apply the squeeze theorem to this inequality. Taking the limit, the right-hand side becomes:
$$\lim_{x \to 0} \frac{q}{x} \frac{x}{p} = 
\lim_{x \to 0} \frac{q}{p}=
\frac{q}{p}$$
and by taking the  limit, the left-hand side becomes:
$$\lim_{x \to 0} \left(\frac{q}{x} - 1 \right) \frac{x}{p} = 
\lim_{x \to 0} \left(\frac{q}{p} - \frac{x}{p}\right) =
\frac{q}{p} - 0 =
\frac{q}{p}$$
Hence $ \lim_{x \to 0} \frac{x}{p} \left \lfloor\frac{q}{x} \right \rfloor =\frac{q}{p}$.
A: Your solution is correct.
Another way to solve it without squeeze theorem: write
$$
\frac{q}{x} = n + r,
$$
where $n \in \mathbb{Z}$ and $0 \le r < 1$. As $x \to 0$, $n \to \infty$ (or $-\infty$ if $q$ is negative) and $r$ varies between $0$ and $1$. Solving for $x$ in the above we have
$$
x = \frac{q}{n + r}.
$$
Now, in the original expression we get
\begin{align*}
\frac{x}{p} \left\lfloor \frac{q}{x} \right\rfloor 
&= \frac{1}{p} \left(\frac{q}{n + r} \right) \lfloor n + r \rfloor \\
&= \frac{1}{p} \left(\frac{q}{n + r} \right) \cdot n \\
&= \frac{q}{p} \cdot \frac{n}{n + r}.
\end{align*}
As $x \to 0$, as we observed earlier $n \to \infty$ and $r$ varies between $0$ and $1$. But as $n \to \infty$, $\frac{n}{n + r} \to 1$ uniformly in $r$ -- meaning that it converges to $1$ regardless of how $r$ varies. So the limit of the above as $x \to 0$ becomes
$$
\frac{q}{p} \cdot 1 = \frac{q}{p}.
$$
A: $$\frac{x}{p}\left\lfloor\frac{q}{x}\right\rfloor=\frac{x}{p}\frac{q}{x}-\frac{x}{p}\left\{\frac{q}{x}\right\}=\frac qp-\frac{x}{p}\left\{\frac{q}{x}\right\}.$$
The second term tends to $0$ because
$$0\le\frac{x}{p}\left\{\dfrac{q}{x}\right\}<\frac{x}{p}.$$
