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Limits and log

by Mattofix
Tags: limits
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Mattofix
#1
Oct21-07, 04:10 PM
P: 142
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)


...?
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Eighty
#2
Oct21-07, 04:32 PM
P: 53
You won't get a 0 in the denominator. You forgot to divide the 1 by n^2.
Mattofix
#3
Oct21-07, 05:31 PM
P: 142
if i divide the bottom by n^2 i get (1/n + 1/n^4)

Mattofix
#4
Oct21-07, 05:31 PM
P: 142
Limits and log

sorry... (1/n - 1/n^4)^(1/2)
Mattofix
#5
Oct21-07, 05:32 PM
P: 142
damn, sorry again, i mean (2/n - 1/n^4)^(1/2)
Mattofix
#6
Oct21-07, 05:33 PM
P: 142
which becomes 0 as n tends to infinity?
Gib Z
#7
Oct22-07, 04:57 AM
HW Helper
Gib Z's Avatar
P: 3,352
[tex]\lim_{n\to \infty} x_n = \lim_{n\to\infty} \left( \frac{ 1+ \frac{\log n}{n^2} }{ \sqrt{ \frac{2n^3-1}{n^4}}} \right)[/tex]

Consider separately, what is the numerator tending towards? How about the denominator?
sennyk
#8
Oct22-07, 07:08 AM
P: 73
Try rationalizing the denominator first. Then things get much easier.


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