How to calculate the limit that seems very complex.. Someone gives me a limit about trigonometric function and combinatorial numbers.
$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom{n}{1}\cos\frac{2}{n^2}+\binom{n}{2}\cos\frac{3}{n^2}\cdots\binom{n}{n}\cos\frac{n+1}{n^2}}+1\right)^n$
when $n$ is big enough, $\displaystyle 0\leqslant\frac{n+1}{n^2}\leqslant \frac\pi 2$ I tried $\displaystyle \frac{x}{1+x}\leqslant\sin x\leqslant x$ 
Use $\sin x\leqslant x$, I got $I\leqslant e$, But use another inequality I got nothing. 
I don't know the answer is $e$ or not. Who can help me. Thanks.

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 $1$.
