# limits and log

by Mattofix
Tags: limits
 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) ...?
 P: 53 You won't get a 0 in the denominator. You forgot to divide the 1 by n^2.
 P: 143 if i divide the bottom by n^2 i get (1/n + 1/n^4)
P: 143

## limits and log

sorry... (1/n - 1/n^4)^(1/2)
 P: 143 damn, sorry again, i mean (2/n - 1/n^4)^(1/2)
 P: 143 which becomes 0 as n tends to infinity?
 HW Helper P: 3,353 $$\lim_{n\to \infty} x_n = \lim_{n\to\infty} \left( \frac{ 1+ \frac{\log n}{n^2} }{ \sqrt{ \frac{2n^3-1}{n^4}}} \right)$$ Consider separately, what is the numerator tending towards? How about the denominator?
 P: 73 Try rationalizing the denominator first. Then things get much easier.

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