How to determine the limit of this sum? I know that $\lim_{x\to\infty} \dfrac{2x^5\cdot2^x}{3^x} = 0$. But what I can't figure out is how to get that answer. One of the things I tried is $\lim_{x\to\infty} 2x^5 \cdot \lim_{x\to\infty}(\dfrac{2}{3})^x$, but then you'd get $\infty \cdot 0$, and I think that is undefined. What would be a correct way to get $0$?

$$F=\lim_{x\to\infty}\frac{2x^5}{(3/2)^x}$$ which is of the form $\dfrac\infty\infty$
If L'Hospital's rule is allowed,
$$F=\lim_{x\to\infty}\frac{2\cdot5x^4}{(3/2)^x\ln3/2}$$ which is again of the form $\dfrac\infty\infty$
So, we can apply L'Hospital's rule again and so on
