Aloha :)
$$\left(\frac{n^2-4n-21}{n^2-3n-28}\right)^n=\left(\frac{n^2-3n-28-n+7}{n^2-3n-28}\right)^n=\left(1-\frac{n-7}{n^2-3n-28}\right)^n$$$$\quad=\left(1-\frac{n-7}{(n-7)(n+4)}\right)^n=\left(1-\frac{1}{n+4}\right)^n=\frac{\left(1-\frac{1}{n+4}\right)^{n+4}}{\left(1-\frac{1}{n+4}\right)^4}\;\;\to\;\;\frac{e^{-1}}{1}=\frac{1}{e}$$
Beachte: \(e^x=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n\).