Help in evaluating: $\lim_{n \to \infty} \frac{\sum_{k=1}^n\lfloor kx\rfloor}{n^2} $ 
Evaluate the following limit.
$$\lim_{n \to \infty} \cfrac{\sum_{k=1}^n\lfloor kx\rfloor}{n^2} $$

I'm not sure how to proceed from here. Normal methods don't seem to apply here, because of the presence of $\lfloor\cdot\rfloor$ (greatest integer function).
Although, I do get a feeling that Sandwich Theorem may come into play here. Not sure how. May be taking interval of $\lfloor x\rfloor$ as : $x - 1 <\lfloor x\rfloor< x $ .
Any one who can guide me through?
 A: As $x - 1 < [x] < x + 1$, we have
$$\cfrac{ x + 2x + 3x + \cdots + nx - n}{n^2} < \cfrac{[x] + [2x] + [3x] + \dots + [nx] }{n^2} < \cfrac{ x + 2x + 3x + \cdots + nx + n}{n^2} $$
See what to do now?
A: Recall that $y-1 < \lfloor y \rfloor \leq y$. Hence, we have
$$\sum_{k=1}^n (kx-1) < \sum_{k=1}^n \lfloor kx \rfloor \leq \sum_{k=1}^n kx $$
This gives us
$$\dfrac{n(n+1)}{2}x -n < \sum_{k=1}^n \lfloor kx \rfloor \leq \dfrac{n(n+1)}2 x$$
Hence, we have
$$\dfrac{\dfrac{n(n+1)}{2}x -n}{n^2} < \dfrac{\sum_{k=1}^n \lfloor kx \rfloor}{n^2} \leq \dfrac{\dfrac{n(n+1)}2 x}{n^2}$$
Taking the limit as $n \to \infty$, we see that
$$\dfrac{x}2 \leq \lim_{n \to \infty}\dfrac{\sum_{k=1}^n \lfloor kx \rfloor}{n^2} \leq \dfrac{x}2$$
Hence, $$\lim_{n \to \infty}\dfrac{\sum_{k=1}^n \lfloor kx \rfloor}{n^2} = \dfrac{x}2$$
A: you can use the Stolz–Cesàro theorem, that states that if b_n is strictly monotone and $$b_n  \rightarrow \inf$$ if $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n} \rightarrow L$$
then we can say $$\frac{a_n}{b_n} \rightarrow L$$
its easy to prove that $$\frac{[nx]}{n^2-(n-1)^2} \rightarrow  \frac{x}{2}$$
using Sandwich Theorem and then using Stolz–Cesàro theorem we get $$\frac{[x]+[2x].....[nx]}{n^2} \rightarrow \frac{x}{2}$$
