$\lim_{n\to\infty} \frac{2^{100+5n}}{e^{4n-10}}= \lim_{n\to\infty} \frac{2^{100}2^{5n}}{e^{4n}e^{-10}}= 2^{100}e^{10}\lim_{n\to \infty}\frac{e^{-4n}}{2^{-5n}}= 2^{100}e^{10}\frac{\lim_{n\to \infty} e^{-4n}}{\lim_{n\to \infty}2^{-5n}} = 0$
I think it's right, both limits exist for the two exponential functions.