# Thread: how to find the bounds of this siquence..

1. ## how to find the bounds of this siquence..

(n^2+2)^0.5 - (n)^0.5

i thought of doing a limit where n->infinity

but here i get undefined form and even if i whould get some finite limite
it will only be one bound

and i cant do limit n->-infinity because its a sequence must be positive??

2. Originally Posted by transgalactic
(n^2+2)^0.5 - (n)^0.5

i thought of doing a limit where n->infinity

but here i get undefined form and even if i whould get some finite limite
it will only be one bound

and i cant do limit n->-infinity because its a sequence must be positive??
We have the sequence $a_n=\sqrt{n^2+2}-\sqrt{n}$. To find the lower bound define $f(n)=\sqrt{n^2+2}-\sqrt{n}$. A little work shows that $f'\left(\approx.886388\right)=0$ and the the second derivative is positive at that value, therefore it is an relative minimum. So now checking 0 the left-endpoint of the domain we find that $f(.886388)\approx.72755$ is a lower bound. Now we can show that $a_n$ is unbounded above, the reason being that for every $\varepsilon>0$ you can find a number $N$ such that $N\leqslant{n}\implies\varepsilon. Also you can note that as n gets arbitrarily large that $a_n\sim\sqrt{n^2}-\sqrt{n}=n-\sqrt{n}$ which clearly is unbounded above.

3. but its sequence "n" must be positive and whole
you cant input n=0.8
?

4. Originally Posted by transgalactic
but its sequence "n" must be positive and whole
you cant input n=0.8
?
I know that, but think about that if $x$ is a local min/ max then the values around it are closer (if continous) to it. So test the biggest integer after the number I gave you and the smallest before.