Limits and log

1. Oct 21, 2007

Mattofix

Find lim (n to infinity) xn

xn = (n^2 + log n)/(2n^3 - 1)^(1/2)

...?

2. Oct 21, 2007

Eighty

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

3. Oct 21, 2007

Mattofix

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

4. Oct 21, 2007

Mattofix

sorry... (1/n - 1/n^4)^(1/2)

5. Oct 21, 2007

Mattofix

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

6. Oct 21, 2007

Mattofix

which becomes 0 as n tends to infinity?

7. Oct 22, 2007

Gib Z

$$\lim_{n\to \infty} x_n = \lim_{n\to\infty} \left( \frac{ 1+ \frac{\log n}{n^2} }{ \sqrt{ \frac{2n^3-1}{n^4}}} \right)$$

Consider separately, what is the numerator tending towards? How about the denominator?

Last edited: Oct 22, 2007
8. Oct 22, 2007

sennyk

Rationalize the denominator

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