Find the Limit as n Approaches Infinity

[Mattofix]

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)


...?

 

[Eighty]

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

 

[Mattofix]

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

 

[Mattofix]

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

 

[Mattofix]

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

 

[Mattofix]

which becomes 0 as n tends to infinity?

 

[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?

 

[sennyk]

Rationalize the denominator

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