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I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$ I know it is $1$ but I don't understand why this is wrong :

$\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$.

So $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}) = \lim \limits_{n\rightarrow\infty}\sin^2(\pi*n\sqrt{1 + \frac{1}{n}}) = \lim \limits_{n\rightarrow\infty} \sin^2(\pi*n) = 0$.

Thanks.

Martin Sleziak
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Desura
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    It looks like you tried to take a limit of just part of the expression. I don't think that's permissible except in a few particular cases, like splitting one limit of a sum into a sum of two limits. – A. Thomas Yerger Dec 22 '15 at 21:29
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    Hint: The function $\sin^2x$ has period $\pi$, i.e. for all real numbers $x$ and all integers $n$ we have $$\sin^2(x+n\pi)=\sin^2x.$$ So for all $n$ you have $$\sin^2(\pi\sqrt{n^2+n})=\sin^2(\pi[\sqrt{n^2+n}-n]).$$ – Jyrki Lahtonen Dec 22 '15 at 21:42
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    And what Alfred Yerger said. In general you should not let $n\to\infty$ in only one part of the formula. Sometimes we can do so, but that's why we prove those simple and "obvious" looking formulas about limits of, say, sums or products of converging sequences. I try to explain this with more words here, but I'm not sure I'm doing a very good job there. – Jyrki Lahtonen Dec 22 '15 at 21:54
  • Yes, I know those theorems, but I think I'm misled by polynomial limits. For example : $$\lim \limits_{n\rightarrow\infty} \frac{2n^3 + n^2 + 1}{n^2 + n + 2} = \lim \limits_{n\rightarrow\infty} = \lim \limits_{n\rightarrow\infty} \frac{n^3(2 + 1/n + 1/n^3)}{n^2(1+ 1/n + 2/n^2)} = \lim \limits_{n\rightarrow\infty} \frac{2n^3}{n^2} = \infty$$ Here, I split the limit into two limits even if one diverges. I know that the right result can be given by Bernouilli L'Hospital, but still. Why is it ok there but not with the square root ? – Desura Dec 22 '15 at 22:10
  • This question is, to some extent related: http://math.stackexchange.com/questions/45759/find-the-limit-lim-limits-n-to-infty-cos-left-pi-sqrtn2-n-right – Martin Sleziak Jan 19 '16 at 12:26

2 Answers2

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First of all, notice that the function $f (\theta) = \sin^2 \theta$ is periodic with a period $\pi$. Then, for any integer $n$ we should have $\sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \left (\pi \left(\sqrt {n^2 + n} - n\right)\right)$. Since $$\lim_{n \to \infty} \left(\sqrt {n^2 + n} - n\right) = \frac {1} {2},$$ it becomes obvious that $$\lim_{n \to \infty} \sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \frac {\pi} {2} = 1.$$

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$$\sqrt{n^2+n} = n \sqrt{1 + \frac{1}{n} } = n (1 + \frac{1}{2n} + O(1/n^2)) =$$ $$ = n + \frac{1}{2} + O(1/n) $$

mlu
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