how should this be resolved?

[jiihaa321]

 

[afos]

L'Hospital's rule.

 

[topsquark]

L'Hospital's rule.

L'Hopital's rule isn't going to help here.
[math]\lim_{x \to \infty} \dfrac{x e^x ln(x)}{e^{x^2}} = \lim_{x \to \infty} \dfrac{(x + 1) ln(x) + 1) e^x }{2 x e^{x^2} } = \lim_{x \to \infty } \dfrac{ (2x + (x + 2) x ~ ln(x) + 1) e^x }{ 2 (2x^2 + 1) e^{x^2} }[/math]
We simply don't have the correct cancellations to make the rule useful.

This is the best I can do. It's probably not the most elegant way. (And it might even be wrong. I need a Math person to check the steps.)

For the sake of clarity I'll do this in two steps.
1) For sufficiently large x, [math]ln(x) < x[/math]. So we can say that
[math]\lim_{x \to \infty } \dfrac{x e^x ln(x)}{e^{x^2}} < \lim_{x \to \infty } \dfrac{x^2 e^x }{ e^{x^2} }[/math]
2) For sufficiently large x, [math]x^2 < e^x[/math]. So we can say that
[math]\lim_{x \to \infty } \dfrac{x^2 e^x }{ e^{x^2} } < \lim_{x \to \infty } \dfrac{e^{2x} }{ e^{x^2} } = \lim_{x \to \infty } \dfrac{1}{ e^{x^2 - 2x} } =0[/math].

Stringing these together:
[math]\lim_{x \to \infty } \dfrac{x e^x ln(x)}{e^{x^2}} < \lim_{x \to \infty } \dfrac{x^2 e^x }{ e^{x^2} } < \lim_{x \to \infty } \dfrac{e^{2x} }{ e^{x^2} } = 0[/math]
So finally
[math]\lim_{x \to \infty } \dfrac{x e^x ln(x)}{e^{x^2}} = 0[/math]
-Dan

 

[Steven G]

You do have some minor issues in my opinion.

In the next to the last last line you have the desired limit being less than 0, yet your last line say that the desired limit is 0.

Also you never stated that the desired limit was not negative.

Other than that very nicely done.