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?

Another way is, because $x$ is approaching $0^+$, i.e. it becomes very small and is positive, then $\exists n\in \mathbb{N}$ s.t. $n\leq \frac{1}{x}<n+1$ or $n=\left\lfloor \frac{1}{x} \right\rfloor$ then $\forall i\in\mathbb{N}$:
$$i\cdot n \leq \frac{i}{x}<i\cdot(n+1) \Rightarrow i\cdot n \leq \left\lfloor \frac{i}{x} \right\rfloor<i\cdot(n+1) \tag{1}$$
and
$$n\left(\sum\limits_{i=1}^k i\right)\leq\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor <(n+1)\left(\sum\limits_{i=1}^k i\right) \iff \\
n\cdot \frac{k(k+1)}{2}\leq\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor <(n+1)\cdot \frac{k(k+1)}{2} \tag{2}$$
but $n\leq \frac{1}{x}<n+1 \Rightarrow \frac{1}{n}\geq x \geq \frac{1}{n+1}$
and $(2)$ becomes 
$$ \frac{n}{n+1}\cdot \frac{k(k+1)}{2} <
x\cdot n\cdot \frac{k(k+1)}{2}\leq \\
x\left(\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor\right) < \\
x\cdot(n+1)\cdot \frac{k(k+1)}{2} \leq
\frac{(n+1)}{n}\cdot \frac{k(k+1)}{2}$$
or 
$$\frac{n}{n+1}\cdot \frac{k(k+1)}{2} < 
x\left(\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor\right) <
\frac{(n+1)}{n}\cdot \frac{k(k+1)}{2} \tag{3}$$
Now, $\lim\limits_{x\rightarrow0^+} \equiv \lim\limits_{n\rightarrow\infty}$ and the result follows from $(3)$.
