Register to reply

Limits and log

by Mattofix
Tags: limits
Share this thread:
Oct21-07, 04:10 PM
P: 141
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)

Phys.Org News Partner Science news on
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
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.
Oct21-07, 05:31 PM
P: 141
if i divide the bottom by n^2 i get (1/n + 1/n^4)

Oct21-07, 05:31 PM
P: 141
Limits and log

sorry... (1/n - 1/n^4)^(1/2)
Oct21-07, 05:32 PM
P: 141
damn, sorry again, i mean (2/n - 1/n^4)^(1/2)
Oct21-07, 05:33 PM
P: 141
which becomes 0 as n tends to infinity?
Gib Z
Oct22-07, 04:57 AM
HW Helper
Gib Z's Avatar
P: 3,348
[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?
Oct22-07, 07:08 AM
P: 73
Try rationalizing the denominator first. Then things get much easier.

Register to reply

Related Discussions
Limits, yes another limits thread. General Math 25
Limits - Calculus & Beyond Homework 3
Limits of MD General Engineering 0
Need to get rid of the 'h' on the denominator Introductory Physics Homework 25