The task is to show the following limit exists, and then compute it. Here, $\,\mathrm{f}:\left[0,1\right] \to \mathbb{R}$ is a continuously differentiable function. $$ \lim_{n \to \infty}\left\{n\left[% \sum_{k = 0}^{n}\,\mathrm{f}\left(k \over n\right) - n\int_{0}^{1}\,\mathrm{f}\left(x\right)\,\mathrm{d}x\right]\right\} $$ It seems so innocent, it should be easy for me at this point$\ldots$ Could anyone provide some direction or intuition on how to approach such problems ?. Specific hints would be helpful as well.
EDIT: The comments below discuss why this limit does not exist in general as written. Perhaps this problem $\left(~\mbox{found on a qualifying exam}~\right)$ had a typo.
We will make some extra assumption on $f(x)$ and shows that under such assumption, the limit diverges unless $f(0)+f(1) = 0$.
Let $f : [0,1] \to \mathbb{R}$ be any $C^2$ function on $[0,1]$, i.e. twice differentiable and the $2^{nd}$ derivative $f''(x)$ is continuous. For any $0 \le a < b \le 1$, let $h = b - a$ and consider following integral
$$\mathcal{I} \stackrel{def}{=} \int_a^b f''(x) (x-a)(b-x) dx$$ Integrate by parts, it is easy to see
$$\begin{align} \mathcal{I} &= \int_a^b (x-a)(b-x) df'(x) = \left[(x-a)(b-x)f'(x)\right]_a^b - \int_a^b f'(x)(a+b - 2x) dx\\ &= \int_a^b (2x - (a+b)) df(x) = \left[(2x-(a+b))f(x)\right]_a^b - 2\int_a^b f(x) dx\\ &= h(f(a)+f(b)) - 2\int_a^b f(x)dx \end{align} $$ Notice the factor $(x-a)(b-x)$ is non-negative on $[a,b]$ with $\displaystyle\;\int_a^b (x-a)(b-x) dx = \frac{h^3}{6}\;$.
If $M$ and $m$ is the maximum and mimumum of $f''(x)$ over $[a,b]$, we have following bound for $\mathcal{I}$.
$$m\frac{h^3}{6} \le \mathcal{I} \le M\frac{h^3}{6}$$
Since $f''(x)$ is continuous over $[a,b]$, by IVT, there exists an $c \in (a,b)$ such that $$\mathcal{I} = f''(c)\frac{h^3}{6}$$
Now for any fixed $n > 1$, sub divide $[0,1]$ into $n$ subintervals $[a_k,b_k] = \left[\frac{k}{n},\frac{k+1}{n}\right]$ for $0 \le k < n$. Apply result above, we find there are $c_k \in [a_k,b_k]$ such that
$$\frac{1}{12n^3} f''(c_k) = \frac{1}{2n}\left(f(a_k)+f(b_k)\right) - \int_{a_k}^{b_k} f(x)dx$$ Summing over $k$ give us
$$n\left\{ \sum_{k=0}^{n-1} \frac12\left[ f\left(\frac{k}{n}\right) +f\left(\frac{k+1}{n}\right) \right] - n\int_0^1 f(x)dx \right\} = \frac{1}{12n}\sum_{k=0}^{n-1}f''(c_k)$$
The RHS has the form of a Riemann sum over $f''(x)$. This means as $n \to \infty$, the limit of LHS exists and equals to $$\lim_{n\to\infty} n\left\{ \sum_{k=0}^{n-1} \frac12\left[ f\left(\frac{k}{n}\right) +f\left(\frac{k+1}{n}\right) \right] - n\int_0^1 f(x)dx \right\}\\ = \frac{1}{12} \int_0^1 f''(x)dx = \frac{1}{12} (f'(1)-f'(0))\tag{*1} $$ Compare this with the sequence in question, we find
$$n\left\{\sum_{k=0}^n f\left(\frac{k}{n}\right) - n\int_0^1 f(x)dx\right\} = \frac{n}{2}(f(0)+f(1)) + \frac{1}{12}(f'(1)-f'(0)) + o(1) $$ From this, we find unless $f(0)+f(1) = 0$, the sequence at hand contains a term proportional to $n$ and diverges in general. If $f(0)+f(1) = 0$, the limit do exists and equals to $\frac{1}{12}(f'(1) - f'(0))$.
As mentioned in comment, this is essentially Euler Maclaurin formula of approximating the integral over $[0,1]$.
As a side note, I think the condition $f$ is $C^2$ can be relaxed. However, I don't know the exact condition for $(*1)$ to be true. $C^2$ is simply the condition one can prove what one need directly.