TITLE: $\lim_{n\to \infty}\sum_{j=0}^{[n/2]} \frac{1}{n} f\left( \frac{j}{n}\right)$
QUESTION [0 upvotes]: If f is continuous in [0,1] then $$\lim_{n\to \infty}\sum_{j=0}^{[n/2]} \frac{1}{n} f\left( \frac{j}{n}\right)$$(where [y] is the largest integer less than or equal to y) 
(A) does not exist
(B) exists and is equal to $\frac{1}{2} \int_0^1 f(x)dx$
(C) exists and is equal to $\int_0^1 f(x)dx$
(D) exists and is equal to $\int_0^{\frac{1}{2}} f(x)dx$
My approach:
              Let I =  $\lim_{n\to \infty}\sum_{j=0}^{n} \frac{1}{n} f( \frac{j}{n})$
Then I= $\int_0^1 f(x)dx$ (by integration as the limit of a sum)
The required integral is therefore ans (B) exists and is equal to $\frac{1}{2} \int_0^1 f(x)dx$
This is area under the function $f(x)$ upto $x=\frac{1}{2}$ which is also = $\int_0^{\frac{1}{2}} f(x)dx$
I am confused between the answers (B) and (D)
Please help.Thank you.

REPLY [1 votes]: The correct answer is $D$. As $n\to \infty$ we may very well write limit as Riemann Sum:
$$\lim_{n\to \infty}\sum_{j=0}^{[n/2]} \frac{1}{n} f( \frac{j}{n}) = \int_{0}^{1/2} f(x) dx$$
Important to note that how limits of riemann integral are calculated. The limits of integral are $\lim_{n\to \infty} \frac{j_{first}}{n}$ and $\lim_{n\to \infty} \frac{j_{last}}{n}$. So the limits are $0$ and $1/2$.
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