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
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}
&amp;=\lim_{n\rightarrow\infty} \frac{x-[x]+2x-[2x]+\dots+nx-[nx]}{n^2}\\
&amp;=\lim_{n\rightarrow\infty} \frac{x(1+2+\dots+n)}{n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
&amp;=x\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
&amp;=\frac x2-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
\end{align*}
hint

$$\lim{n\rightarrow \infty} \frac{(x) [1+2+3 +. . . + n]}{n^2}$$
 As$ 1+2+3+. . .+n = \frac{n(n-1)}{2}$
So 
$$\lim{ n \rightarrow \infty} \frac{1-\frac{1}{n}}{1} x$$
 Answer is $x$