L'Hospital's Rule for Limit of x^a ln(x) as x Approaches 0 (a>0)

[fxo]

Homework Statement

Use L'Hospitals rule to show that lim x->0 x^a ln(x) = 0

I don't know how to solve this. I guess the first thing to do is to transform it in some way so that one can use L'Hospitals rule, but I don't know how.

Thank you!

EDIT: a>0

It's not a differential equation as the headline says, my bad.

 

[LCKurtz]

Homework Statement

Use L'Hospitals rule to show that lim_{x\rightarrow 0} x^a ln(x) = 0

I don't know how to solve this. I guess the first thing to do is to transform it in some way so that one can use L'Hospitals rule, but I don't know how.

Thank you!

Is there something else you haven't told us? Like $a>0$? Write it like this:$$
\lim_{x\rightarrow 0}\frac {\ln x}{\frac 1 {x^a}}$$

 

[fxo]

Is there something else you haven't told us? Like $a>0$? Write it like this:$$
\lim_{x\rightarrow 0}\frac {\ln x}{\frac 1 {x^a}}$$

Oh sorry you're right I totally missed that a>0. I get it then.

Can I ask a follow-up question based on this one?

By setting x=ln(t) in the equation above(the one I asked for help on first) show that:

$$
\lim_{x\rightarrow -∞} |x|^a e^x = 0$$

This one really bugs my mind I just don't get it.

 

[LCKurtz]

Oh sorry you're right I totally missed that a>0. I get it then.

Can I ask a follow-up question based on this one?

By setting x=ln(t) in the equation above(the one I asked for help on first) show that:

$$
\lim_{x\rightarrow -∞} |x|^a e^x = 0$$

This one really bugs my mind I just don't get it.

You should note that in the original limit, it is really $x\rightarrow 0^+$. What happens if you substitute $x=\ln t$ in the original problem?

 

[fxo]

$$
\lim_{ln(t)\rightarrow 0}\frac {\ln (ln(t))}{\frac 1 {ln(t)^a}}$$

So I get $$
ln(ln(t))ln^a(t)$$

But that expression don't tell me much all I know is that it's root should be e.

 

[LCKurtz]

Woops, I didn't mean that. I meant to try $t=\ln x$ which is $x=e^t$. I expect that is a typo in your text.

 

[fxo]

Woops, I didn't mean that. I meant to try $t=\ln x$ which is $x=e^t$. I expect that is a typo in your text.

$\lim_{e^t\rightarrow 0}\frac {\ln (e^t}{\frac 1 {e^t^a}}$

I don't really get it what's the connection to the new limit, do they want me to change the e^x to x=ln(t) and get |x|^a * t instead?

 

[LCKurtz]

$\lim_{e^t\rightarrow 0}\frac {\ln (e^t)}{\frac 1 {(e^t)^a}}$

C'mon, do the algebra simplification. And express it as$$
\lim_{t\rightarrow\, ?}...$$Then you might see a connection to your new problem.

 

[fxo]

C'mon, do the algebra simplification. And express it as$$
\lim_{t\rightarrow\, ?}...$$Then you might see a connection to your new problem.

By setting $x=e^t$ in the first one I got$$
\lim_{t\rightarrow\, 0} te^t^a $$
I'm starting to see that it's close to the form in b but not exactly, should I use L'Hospitals rule from here to show that as this goes to infinity the limit goes to zero?

 

[LCKurtz]

By setting $x=e^t$ in the first one I got


$$
\lim_{t\rightarrow\, 0} te^t^a $$
I'm starting to see that it's close to the form in b but not exactly, should I use L'Hospitals rule from here to show that as this goes to infinity the limit goes to zero?

Tex won't accept e^t^a with no parentheses. Anyway, do you consider that to be simplified? It is more properly written $(e^t)^a$, but still not simplified until you write it as $e^{at}$. And you have $t\rightarrow 0$. If $x=e^t$ and $x\rightarrow 0^+$, does $t\rightarrow 0$? What should it be?

That being said, you are correct that you can massage it into $|x|e^x$ form but it still isn't in the $|x|^ae^x$ form requested. So I'm afraid my suggestion to correct what I think is a misprint isn't going to finish it. To tell the truth, I don't see why your author wants you to do it that way anyway. It is easy to work it directly. Letting $x = -t$ we have$$
\lim_{x\rightarrow {-\infty}}|x|^ae^x=\lim_{t\rightarrow \infty}t^ae^{-t}
=\lim_{t\rightarrow \infty}\frac{t^a}{e^{t}}$$which you can now do with repeating L'Hospitals rule until the exponent on $t$ goes negative.