MHB Simplifying Limit Problem - L'Hôpital and Algebra

[Lepros]

I have the following limit problem: $$\Large\lim_{x \rightarrow \infty}\frac{x^{\frac{5}{3}}}{e^{2x}}$$

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?

 

[Also sprach Zarathustra]

I have the following limit problem: $$\Large\lim_{x \rightarrow \infty}\frac{x^{\frac{5}{3}}}{e^{2x}}$$

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?

You need use L'Hôpital's rule two times.

First:$$ \Large\lim_{x \rightarrow \infty}\frac{\frac{5}{3}x^{\frac{2}{3}}}{2e^{2x}} $$second:$$\Large\lim_{x \rightarrow \infty}\frac{\frac{5}{3}\frac{2}{3}x^{-\frac{1}{3}}}{2e^{2x}}$$$$\Large\lim_{x \rightarrow \infty}\frac{\frac{5}{3}\frac{2}{3}} 2e^{-2x}x^{-\;\frac{1}{3}}$$And now it's clear that the limit equals to $0$.

 

[tkhunny]

I can tell that the limit is 0

No, you can't. Why did you put x^(2/3) in the denominator? Why not leave it in the numerator and apply the rule again?