Find $\displaystyle \lim_{n\to \infty} \frac{n^5+2\sin n}{e^{-n}+6n^5}.$
the largest that $\displaystyle 2\sin(n)$ can ever be is 2.
$\displaystyle e^{-n} \to 0$ really fast as $\displaystyle n \to \infty$
so, in essense you have a fraction on the order of $\displaystyle \frac{n^5 + k}{6n^5}$ , where $\displaystyle -2 \le k \le 2$
the limit is $\displaystyle \frac{1}{6}$