Evaluate the limit $\lim\limits_{n\to0}\frac{(x)+(2x)+\cdots (nx)}{n^2}$ Find the limit of $\lim_{n\rightarrow ~0}\frac{(x)+(2x)+\cdots (nx)}{n^2}$, where, $(x)=x-[x]$ and $[x] $ is the greatest integer function(the fractional part function).
I feel, as $n \rightarrow 0$ this limit  goes to infinity, but the options given are $x~,~x/2,~x/3,~x/4$.
How this is happening, I double checked the question paper, in question $n$. is tending to 0 only not to $\infty$.
I found a similar question here
 A: This was too long for a comment, hopefully this helps:
Using $1+2+\dots+n=\frac{n(n+1)}{2}$ we find that:
\begin{align*}
\lim_{n\rightarrow\infty} \frac{(x)+(2x)+\dots+(nx)}{n^2}
&=\lim_{n\rightarrow\infty} \frac{x-[x]+2x-[2x]+\dots+nx-[nx]}{n^2}\\
&=\lim_{n\rightarrow\infty} \frac{x(1+2+\dots+n)}{n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
&=x\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
&=\frac x2-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
\end{align*}
A: For $n \longmapsto 0$ the nominator is meaningless!
For $n \longmapsto +\infty$ we have:
$$\forall_y:0 \leqslant y-[y] <1 \Longrightarrow 0 \leqslant (y) <1$$
$$\lim_{n \longmapsto +\infty} \frac{(x)+...+(nx)}{n^2} = \lim_{n \longmapsto +\infty} \frac{(x)+...+(nx)}{n}.\frac{1}{n} =$$
$$= \lim_{n \longmapsto +\infty} \frac{Mean((x),...,(nx))}{n} = \lim_{n \longmapsto +\infty} \frac{Bounded}{n} = 0$$
In fact the mean of (ix)'s is stays in $[0,1)$ since all of them are in that area.
A: not an answer to the stated question  (we use denominator $n$, not $n^2$) but interesting nonetheless  
See information on equidistributed sequences ... 
If $x$ is irrational, then the sequence $(x), (2x), (3x), \dots$ of fractional parts is equidistributed on $[0,1]$.  Therefore, for every Riemann integrable function on $[0,1]$ we have
$$
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n f\big((kn)\big) = \int_0^1 f(t)\;dt .
$$
The function $f(t) = t$ is Riemann integrable, so
$$
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n  (kn) = \int_0^1 t\;dt =\frac{1}{2}.
$$
if $x$ is irrational.  If $x$ is rational, this limit could be something other than $1/2$.
