1rehnde........ it lies between -1 and 1
9 Answers
1subhi is not 'n' an integer?? i think it is...
\lim_{x\rightarrow infinity} cos(n\pi \sqrt{1 +1/n}) = \lim_{x\rightarrow infinity} cos(2n\pi - n\pi \sqrt{1 +1/n}) = \lim_{x\rightarrow infinity} cos{n\pi (\frac{4-1-1/n}{2 + \sqrt{1 + 1/n}}} = \lim_{x\rightarrow infinity} cos{n\pi(-3/2n)} = \lim_{x\rightarrow infinity} cos(-3\pi /2) = 0
1hey listen
if n is real number and n→∞ then graph of \Pi \left(\sqrt{n^{2}+n} \right) is continuous and strictly increasing so cos\left( \Pi \left(\sqrt{n^{2}+n} \right) \right) continuously varies betwean -1 to 1
10 comes only at integral values of n otherwise doesn't exists......