2

Find the limit following:

$$L=\lim_{n\to \infty} (-1)^{n}\sin\left(\pi\sqrt{n^2+n}\right)$$

Thanks in advance!

Norbert
  • 56,803
Iloveyou
  • 2,503
  • 1
  • 16
  • 18

1 Answers1

3

$$\sin{(x+k\pi)}=(-1)^k\sin{x}$$ so $$(-1)^n\sin{\pi\sqrt{n^2+n}}=(-1)^n\sin{(\pi(\sqrt{n^2+n}-n)+n\pi)}=\sin{\dfrac{n\pi}{\sqrt{n^2+n}+n}}\to 1,n\to \infty$$

math110
  • 93,304
  • $L\neq 1$. $(-1)^n\sin(\pi(\sqrt{n^2+n}-n)+n\pi)\neq (-1)^n\sin(\pi(\sqrt{n^2+n}-n))\neq \sin\frac{n\pi}{\sqrt{n^2+n}+n}$ – Iloveyou Nov 23 '13 at 07:16
  • $(-1)^n\cdot(-1)^n=?$ – math110 Nov 23 '13 at 07:20
  • Result: Limit does not exist – Iloveyou Nov 23 '13 at 07:31
  • @math110, nice approach. I've a query, the first equality is defined for integer $n,$ right? – lab bhattacharjee Nov 23 '13 at 07:45
  • 1
    @labbhattacharjee,Thank you, I think this problem $n$ is integer,because in china book,Our rules in limit $n$ mean integer. – math110 Nov 23 '13 at 07:55
  • @math110: You try to use Toylor's expansion. And the other is a result of your result. – Iloveyou Nov 23 '13 at 08:03
  • In Indian and USA books too, $n \to \infty$ is supposed to mean tending to $\infty$ through integer values unless otherwise stated. The limit is $1$ under this condition and does not exist otherwise because if $n$ is supposed to be a real number tending to $\infty$ then we can't define $(-1)^{n}$ in the real number system. – Paramanand Singh Nov 23 '13 at 18:29