L'Hospital's rule with infinity (limits w/ differentiation)

[Melissaherman]

Alright, here is the problem I'm working on:

I'm trying to find the Limit as x---> infinity of:[(ln x)squared]/x
and am supposed to use l'hospitals rule, because the limit on its own would be infinity/infinity.
So my understanding of L'Hospitals rule is that if the limit equals something like 0/0 or infinity/infinity, you just differentiate the top and bottom, and that's it...
So the denominator becomes 1 after differentiating; for the top then I ended up with [2(lnx)]/x (from using the chain rule and differentiating the squared function and leaving lnx, and then differentiating lnx to get 1/x)
Which still equals infinity/infinity, not??
So now I don't know if I've just done something wrong, or if this is just the answer... but that doesn't seem right to me.
Thanks for any help

 

[galactus]

You apply L'Hopital again.

After the first time around, you ended up with:

\(\displaystyle \L\\\frac{2ln(x)}{x}\), correct?.

Use L'Hopital again using this.

You get 2/x. That's what you got.

Well, what's the limit as x approaches infinity?. You have it.

 

[Melissaherman]

ooooh ok. I didnt think to use the rule again... that would make sense, haha. Thanks for the help, I really appreciate it!
Melissa