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Calculate

$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$

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Is $n$ here required to be an integer, or a real number? –  MJD Apr 2 '13 at 22:10
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As MJD hints, if n is a real number, then your limit doesn't exist, since you're bouncing between 0 and 1. –  Calvin Lin Apr 2 '13 at 22:14
    
The answers all take $n$ to be restricted to the integers, their answers would equally apply if $n$ was restricted to the naturals as well, right? –  user66807 Apr 23 '13 at 0:20
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5 Answers

Edit: This assumes that $n$ is restricted to the integers.

Firstly, since $ \mid \sin \theta \mid $ is periodic with period $\pi$, we want to look at the value of $ \pi \sqrt{n^2 + n +1} \pmod{\pi}$.

Secondly, convince yourself that $\lim_{n\rightarrow \infty} \pi \sqrt{n^2+n+1} \pmod{\pi} = \frac {1}{2} \pi $.

Thirdly, by the continuity of $\sin \theta$, $ \mid \lim_{n \rightarrow \infty} \sin (\pi \sqrt{n^2+n+1} ) \mid = \mid \sin \lim_{n\rightarrow \infty} \pi \sqrt{n^2+n+1}\mid = \sin \frac {1}{2} \pi$.

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Can you help us convince ourselves? –  levitopher Apr 2 '13 at 22:16
    
So, it's oscillating around $\pm1$, no? –  Berci Apr 2 '13 at 22:17
    
Your answer is correct! But, I did: $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|=\lim \limits_{n \to \infty} \left |\sin\pi \cdot n \sqrt{\dfrac{1}{n^2}+\dfrac{1}{n}+1} \right |$. But $\lim \limits_{n \to \infty} \sqrt{\dfrac{1}{n^2}+\dfrac{1}{n}+1}=1$. So $$\lim \limits_{n \to \infty} \left |\sin\pi \cdot n \sqrt{\dfrac{1}{n^2}+\dfrac{1}{n}+1} \right |=\left |\sin \left ( \lim \limits_{n \to \infty} \pi \cdot n \sqrt{\dfrac{1}{n^2}+\dfrac{1}{n}+1} \right ) \right |=0$$ why is this wrong? –  lechuza Apr 2 '13 at 22:19
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Because of the multiplying by $n$, to the front of the square root. You should be approximating your square root by $1 + \frac {1}{2n}$ instead. Moreover, note that $lim_{n\rightarrow \infty} n f(n) \neq n \lim_{n \rightarrow \infty} f(n)$. –  Calvin Lin Apr 2 '13 at 22:24
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Note that $$\sqrt{n^2+n+1}-n=\frac{n+1}{\sqrt{n^2+n+1}+n}\to 1/2$$ as $n\to\infty$.

For even $n$, $\sin(\sqrt{n^2+n+1}\pi)=\sin(\sqrt{n^2+n+1}\pi-n\pi)\to \sin(\pi/2)=1$ as $n\to\infty, n$ even.

For odd $n$, $\sin(\sqrt{n^2+n+1}\pi)=-\sin(\sqrt{n^2+n+1}\pi-n\pi)\to -\sin(\pi/2)=-1$ as $n\to\infty, n$ odd.

Therefore, $$|\sin(\sqrt{n^2+n+1}\pi)|\to 1$$ as $n\to\infty$.

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up vote 4 down vote accepted

The function $|\sin x|$ is periodic with period $\pi$. Hence

$$\lim \limits_{n \to \infty}|\sin \pi \sqrt{n^2+n+1}|= \lim \limits_{n \to \infty}|\sin \pi (\sqrt{n^2+n+1}-n)|$$

But

$$\lim \limits_{n \to \infty} (\sqrt{n^2+n+1}-n)=\lim \limits_{n \to \infty} \dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\dfrac{1}{2}$$

It follows that the limit we are computing is equal to $\left |\sin \dfrac{\pi}{2}\right |$, which is 1.

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Yes, that's a neat elementary way. But don't avoid Taylor too long...One day you'll have no choice! –  1015 Apr 2 '13 at 22:35
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If you know a little bit about Taylor expansions and big O notation, you can do that easily. We will use $$ \sqrt{1+u}=1+\frac{u}{2}+O(u^2)\qquad \mbox{when } u\longrightarrow 0. $$ Now $$ \pi\sqrt{n^2+n+1}=\pi n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=\pi n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)=\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right) $$ So $$ \sin (\pi\sqrt{n^2+n+1})=\sin\left(\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)=(-1)^n\sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right). $$ It follows that $$ |\sin (\pi\sqrt{n^2+n+1})|=\left| \sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)\right|\longrightarrow \left| \sin\left(\frac{\pi}{2}\right)\right|=1. $$

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We have $$\sqrt{n^2+n+1}=n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=n\left(1+\frac{1}{2n}+o(\frac{1}{n})\right)=n+\frac{1}{2}+o(1)$$ hence $$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|=\lim \limits_{n \to \infty}|\sin(n\pi+\frac{\pi}{2}+o(1))|=\lim \limits_{n \to \infty}|\cos(o(1))|=1.$$

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