Calculus
posted by Jasmine .
Another one I'm not sure about is the limit as x approaches infinity of (x^2)/lnx
I took the derivative of the top and bottom and got 2x/(1/x) which is the same thing as 2x^2. Can I keep going or do I just say that the limit does not exist?

you can keep going, but if you do, you're in trouble.
2x^2 = 2x^2/1
using the Rule, you'd have
4x/0 > oo
So, the limit is as you found it, 2x^2 > oo