21is the answer 1?
7 Answers
1708$\boldsymbol{Ans:}\Rightarrow $\lim_{n \to \infty }cos(\pi.\sqrt{n^2+n})=\lim_{n \to \infty}(-1)^n.cos(n\pi-\pi.\sqrt{n^2+n})$\\\\ for $n$ is even as well as odd positive Integer. Means $n\epsilon2m$ OR $n\epsilon2m+1$\\\\ for $m\epsilon Z^+$ \\\\ =$\lim_{n \to \infty}(-1)^n.cos\left \{\frac{(n\pi-\pi\sqrt{n^2+n})\times(n\pi+\pi\sqrt{n^2+n})}{n\pi+\pi\sqrt{n^2+n}} \right \}$\\\\ =$\lim_{n \to \infty}(-1)^n.cos\left \{ \frac{n^2\pi^2-n^2\pi^2-n\pi^2}{n\pi+\pi\sqrt{n^2+n}}\right \}$\\\\ =$\lim_{n \to \infty}(-1)^n.cos\left \{ \frac{-n\pi^2}{n\pi+\pi\sqrt{n^2+n}} \right \}$\\\\ =$(-1)^ncos(\frac{-\pi}{2}) = 0$
1are two n's used here same.....
bcoz if they are ...one is integer and other is not....
---i dont know..ne intuition....
1yes if n is an integer then....
..i can get some intuition from the fact that as n tends inf. n2 is much more than n so root(n2+n) will be approximated to n and then
sin(pi*n) is 0 .......
omg...its cos here :P...i m still stuck..