Homework Help: Calculus

Posted by Jasmine on Friday, March 30, 2012 at 12:37pm.

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?

Calculus - Steve, Friday, March 30, 2012 at 2:01pm
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

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