Find $\lim\limits_{x\to0^+}x(\lfloor \frac{1}{x}\rfloor+\lfloor \frac{2}{x}\rfloor+\cdots+\lfloor \frac{k}{x}\rfloor), \, k \in \mathbb N$. $$M:=x\left(\left\lfloor \frac{1}{x} \right\rfloor+\left\lfloor \frac{2}{x}\right\rfloor+\cdots+\left\lfloor \frac{k}{x}\right\rfloor\right),\, k \in \mathbb N.$$
Using $\lfloor y \rfloor=y-\{y\}$, 
$$M=x\sum_{i=1}^{k}\left(\frac{i}{x}-\left\{\frac{i}{x}\right\}\right)=\frac{k(k+1)}{2}-x \sum_{i=1}^{k}\left\{\frac{i}{x}\right\}$$ 
Since $0\leq\{y\}<1$, the coefficient of $x$ is finite and thus, $$\lim_{x\to0^+}M=\frac{k(k+1)}{2}$$
Is this correct?

Your proof is correct. Here is another way to compute this limit.
Note that $\frac{i}{x}-1 < \left \lfloor \frac{i}{x} \right \rfloor \le \frac{i}{x}$,$\forall i=\overline{1,k}$.
Therefore, $\frac{k(k+1)}{2x} -k < \left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \frac{k}{x} \right \rfloor \le \frac{k(k+1)}{2x}$.
Hence, $\frac{k(k+1)}{2} -kx < x\left(\left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \frac{k}{x} \right \rfloor \right) \le \frac{k(k+1)}{2}$.
By the squeeze theorem we get that your limit equals $\frac{k(k+1)}{2}$ as you proved.
