$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n\left[\sqrt{4i/n}\right]$ $$
\lim_{n \to \infty}
\sum_{i = 1}^{n}{1 \over n}
\,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor\quad
\mbox{where}\ \left\lfloor\,x\,\right\rfloor\
\mbox{is the greatest integer function.}
$$
I approached the problem this way- 
$$
\lim_{n \to \infty}
\sum_{i = 1}^{n}{1 \over n}
\,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor
=
\lim_{n \to \infty}\frac{1}{n}\int_{n/4}^{n}\mathrm{d}x =
\lim_{n \to \infty}\left(\frac{1}{n}\times\frac{3n}{4}\right) =
\frac{3}{4}
$$

I felt like a bad-ass doing this, only to find that the answer is actually $3$. Where did I go wrong? How do I make it right ?.

The Riemann integral is $$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + \frac{b-a}{n} k\right) = \int_{x=a}^b f(x) \, dx,$$ and with $a = 0$, $b = 1$,and $f(x) = \lfloor 2 \sqrt{x} \rfloor$, we obtain $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \lfloor 2 \sqrt{k/n} \rfloor = \int_{x=0}^1 \lfloor 2 \sqrt{x} \rfloor \, dx.$$  Since $2 \sqrt{x} = 1$ when $1/4 \le x < 1$, and $0$ if $0 \le x < 1/4$, this becomes $$\int_{x=1/4}^1 \, dx = \frac{3}{4}.$$  It is not possible for the given sum to equal $3$, since it is trivially bounded above by $$\frac{1}{n} \sum_{k=1}^n \left\lfloor 2\sqrt{\frac{k}{n}} \right\rfloor < \frac{1}{n} \sum_{k=1}^n 2 \sqrt{\frac{k}{n}} < \frac{1}{n} \sum_{k=1}^n 2 = \frac{2n}{n} = 2.$$
