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Simplifying a Limit

Lepros

New member
Feb 7, 2012
3
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?
 
Jan 31, 2012
54
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 $0$.
 
Last edited by a moderator:

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
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?