Sequence Question

[uberathlete]

Hi everyone. I'm kinda stuck on how to find the limit of this sequence as n approaches infinity:

a = [sqrt(4n^2 + 3)] / [(n - sqrt(n)]

I'm not really sure how to begin. If someone can point me in the right direction it would greatly be appreciated. thanks!

 

[Daniel_Feldman]

Hi everyone. I'm kinda stuck on how to find the limit of this sequence as n approaches infinity:

a = [sqrt(4n^2 + 3)] / [(n - sqrt(n)]

I'm not really sure how to begin. If someone can point me in the right direction it would greatly be appreciated. thanks!

You need the highest degree of the numerator and to compare it to the highest degree of the denominator. In the denominator you have n^1 and n^(1/2), so n^1? In the numerator you have root(4n^2)=2n. Take the limit as n approaches infinity of 2n/n and you get...?

 

[uberathlete]

Hi everyone. I'm kinda stuck on how to find the limit of this sequence as n approaches infinity:

a = [sqrt(4n^2 + 3)] / [(n - sqrt(n)]

I'm not really sure how to begin. If someone can point me in the right direction it would greatly be appreciated. thanks!

You need the highest degree of the numerator and to compare it to the highest degree of the denominator. In the denominator you have n^1 and n^(1/2), so n^1? In the numerator you have root(4n^2)=2n. Take the limit as n approaches infinity of 2n/n and you get...?

Hi Daniel. I think you misread the numerator. It's root(4n^2 + 3), not root(4n^2).

 

[Gene]

A semi-valid criticism.
Factor 4x^2 from the numerator.
[sqrt(4n^2 + 3)] / [(n - sqrt(n)] =
2n*sqrt[1+3/(4n^2)]/[(n - sqrt(n)] =
2*sqrt(1+(3/4n^2)/{1-1/sqrt(n)
As n -> oo that becomes
2*1/1

 

[uberathlete]

A semi-valid criticism.
Factor 4x^2 from the numerator.
[sqrt(4n^2 + 3)] / [(n - sqrt(n)] =
2n*sqrt[1+3/(4n^2)]/[(n - sqrt(n)] =
2*sqrt(1+(3/4n^2)/{1-1/sqrt(n)
As n -> oo that becomes
2*1/1

Oh man, I totally didn't see that. :x Dang. Thanks Gene.