
#1
Oct2107, 04:10 PM

P: 143

Please help, i take out n^2 top and bottom so end up with 0 as demominator....
Find lim (n to infinity) xn xn = (n^2 + log n)/(2n^3  1)^(1/2) ...? 



#2
Oct2107, 04:32 PM

P: 53

You won't get a 0 in the denominator. You forgot to divide the 1 by n^2.




#3
Oct2107, 05:31 PM

P: 143

if i divide the bottom by n^2 i get (1/n + 1/n^4)




#4
Oct2107, 05:31 PM

P: 143

limits and log
sorry... (1/n  1/n^4)^(1/2)




#5
Oct2107, 05:32 PM

P: 143

damn, sorry again, i mean (2/n  1/n^4)^(1/2)




#6
Oct2107, 05:33 PM

P: 143

which becomes 0 as n tends to infinity?




#7
Oct2207, 04:57 AM

HW Helper
P: 3,353

[tex]\lim_{n\to \infty} x_n = \lim_{n\to\infty} \left( \frac{ 1+ \frac{\log n}{n^2} }{ \sqrt{ \frac{2n^31}{n^4}}} \right)[/tex]
Consider separately, what is the numerator tending towards? How about the denominator? 



#8
Oct2207, 07:08 AM

P: 73

Try rationalizing the denominator first. Then things get much easier.



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