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I have reduced it to the following using L'Hôpital's rule once and basic algebra: $$\Large\lim_{x \rightarrow \infty}\frac{5}{2e^{2x}3x^{\frac{-2}{3}}}$$

I can tell that the limit is 0, but I'm wondering if I could reduce this further to make that even more apparent, or approach it in another manner that simplifies it better?