(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??
(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 $\displaystyle a_n=\sqrt{n^2+2}-\sqrt{n}$. To find the lower bound define $\displaystyle f(n)=\sqrt{n^2+2}-\sqrt{n}$. A little work shows that $\displaystyle f'\left(\approx.886388\right)=0$ 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 $\displaystyle f(.886388)\approx.72755$ is a lower bound. Now we can show that $\displaystyle a_n$ is unbounded above, the reason being that for every $\displaystyle \varepsilon>0$ you can find a number $\displaystyle N$ such that $\displaystyle N\leqslant{n}\implies\varepsilon<a_n$. Also you can note that as n gets arbitrarily large that $\displaystyle a_n\sim\sqrt{n^2}-\sqrt{n}=n-\sqrt{n}$ which clearly is unbounded above.