Wednesday

April 1, 2015

April 1, 2015

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

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 2:01pmyou 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

**Answer this Question**

**Related Questions**

Calculus - Find the limit as x approaches infinity of (lnx)^(1/x). This unit is ...

calculus - The limit as x approaches infinity of (e^x+x)^(1/x). I got that it ...

calculus - (a) find the intervals on which f is incrs or decrs. (b) find the ...

BC Calculus - lim x ->infinity (e^x/lnx) I got infinity/0 but is that allowed...

Calculus - So this one is confusing me: limit of n as it approaches infinity of...

Calculus - f(x) = (ax+b)/(x^2 - c) i) the graph of f is symmetric about the y-...

Calculus - Find limit as x approaches 1 5/(x-1)^2 A. 0 B. Negative infinity C. 5...

Differential Calculus - use the rule that says limit of (e^h - 1)/h = 1 as h ...

Differential Calculus - use the rule that says limit of (e^h - 1)/h = 1 as h ...

calculus - use the rule that says limit of (e^h - 1)/h = 1 as h approaches 0 to ...