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
Respond to this Question
Similar Questions

Calculus
f(x) = (ax+b)/(x^2  c) i) the graph of f is symmetric about the yaxis ii) limit as x approaches 2+ of f(x) is positive infinity iii) f'(1) = 2 Determine the values of a, b, c I got that c = 4 from the first i). I'm stuck on the … 
calculus
what is the answer for the integral of (1/(xln(x)) from 1 to infinity? 
calculus
(a) find the intervals on which f is incrs or decrs. (b) find the local min/max values of f. (c) find the intervals of concavity and the inflection points. f(x)=(lnx)/sqrtx I took the derivative and got 2lnx/2x^3/2 take the values … 
BC Calculus
lim x >infinity (e^x/lnx) I got infinity/0 but is that allowed to be my answer? 
calculus
The limit as x approaches infinity of (e^x+x)^(1/x). I got that it diverges, but I'm not sure if I made a mistake. My work: lim(e+x^(1/x)) lim(e+(1/x^x)) lim(ex^x+1)/x^x l'hopital:lim(e^(e(lnx+1))+1)/e^(lnx+1) diverges? 
Calculus
Find the limit as x approaches infinity of (lnx)^(1/x). This unit is on L'Hopital's rule. I know that the answer is 1, I just don't know how to get there. I tried taking the ln of everything so that you have ln(the whole limit) = limx>infinity … 
MathCalculus
Hi, I am trying to figure out what the limit as h approaches 0 of (12h)^(1/h) is. I am unfamiliar with the process I am supposed to use to solve this limit. I have just been reasoning out this limit, but I keep getting the answer, … 
Calculus
Find limit as x approaches 1 5/(x1)^2 A. 0 B. Negative infinity C. 5/4 D. Infinity If I use limit as h approaches 0 f(x+h)f(x)/h , will I get an x in the answer? 
Calculus
So this one is confusing me: limit of n as it approaches infinity of (n{SIGMA}i=1) (1/(2^i) *the n is on top of the sigma notation and i=1 is at the bottom. I understand that 'i' would equal (n(n+1))/2 however the algebra from there … 
Calculus
what is the limit as h approaches 0 of ((x+h)^pix^pi)/h?