How to calculate the limit that seems very complex..

HINT:

$$C+i\cdot S=\sum_{r=0}^n\left[\binom nr\left(\cos\dfrac{r+1}{n^2}+i\sin\dfrac{r+1}{n^2}\right)\right]$$

$$=\sum_{r=0}^n\binom nre^{i(r+1)/n^2}$$

$$=e^{i/n^2}\sum_{r=0}^n\binom nr(e^{i/n^2})^r$$

$$=e^{i/n^2}(1+e^{i/n^2})^n$$

Now $1+e^{2iy}=2\cos y(\cos y+i\sin y)$

and use de Moivre's Formula

With rough (but safe) approximations, we can avoid the use of trigonometric stuff, that is $$l=\lim_{n\to \infty } \, \left(1+\frac{n+2}{2n^2}\right)^n=\lim_{n\to \infty } \, \exp\left(\frac{ n+2}{2n}\right)=\sqrt{e}$$

where $\displaystyle \sin\left(\frac{k}{n^2}\right)$ behaves like $\displaystyle \frac{k}{n^2}$ and $\displaystyle \cos\left(\frac{k}{n^2}\right)$ behaves like $1$.

Consider $$\lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\cos\frac{k+1}{n^2}}.$$ If this limit equals $\alpha$, your limit equals $e^\alpha$. In order to simplify a bit, thanks to the Taylor expansions of $\sin$ and $\cos$ near $0$, we may multiply top and bottom by $n^2$: $$\begin{align} \lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}n^2\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}n^2\cos\frac{k+1}{n^2}}&=\lim_{n\to\infty} {n}\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}(k+1)}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\left(n^2-\frac{1}{2}\frac{(k+1)^2}{n^2}\right)} \\ &= \lim_{n\to\infty} {n}\frac{\displaystyle\sum\limits_{k=0}^n k{n \choose k} + \displaystyle\sum\limits_{k=0}^n {n \choose k}}{n^2\displaystyle\sum\limits_{k=0}^n {n \choose k} - \frac{1}{2n^2}\displaystyle\sum\limits_{k=0}^n (k+1)^2{n \choose k}} \\ &= \lim_{n\to\infty} \frac{2^{n-1}n(n+2)}{n^2 2^n-2^{n-3}(1+5/n+4/n^2)}=\frac{1}{2}.\end{align}$$ Thus your expression approaches $\sqrt{e}$.