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$$\lim_{n\to\infty }\,n\,\sin\frac{1}{n^{2}+1}$$ How to solve this limit? It seems to be 0. But I don't know how to justify it.
Thanks for your help

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  • $\begingroup$ The last $n$ is outside $\sin$ or inside? $\endgroup$
    – Error 404
    Nov 14, 2016 at 8:23
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    $\begingroup$ @Speedding Then it would be wise, I think, to write that $\;n\;$ before the sine, namely $$n\sin\frac1{n^2+1}$$ $\endgroup$
    – DonAntonio
    Nov 14, 2016 at 8:28

4 Answers 4

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\begin{align} \lim_{n\to\infty }\sin (\frac{1}{n^{2}+1})n&=\lim_{n\to\infty }\frac{\sin (\frac{1}{n^{2}+1})}{\frac1{n^2+1}}\times \frac1{n^2+1}n\\ &=\lim_{n\to\infty }\frac{\sin (\frac{1}{n^{2}+1})}{\frac1{n^2+1}} \times \lim_{n\to\infty } \frac n{n^2+1}\\ &=1 \times 0 \end{align}

Where you may recall that $\lim_{x\to 0}\frac{\sin x}{x}=1$.

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  • $\begingroup$ Wow it seems to be pretty easy and elegant solution. Thank you very much $\endgroup$
    – Speedding
    Nov 14, 2016 at 8:51
  • $\begingroup$ You are welcome! $\endgroup$
    – Math-fun
    Nov 14, 2016 at 9:05
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Use $$ |\sin x|\le\min(|x|,1). $$ to find $$\frac{n}{1+n^2}$$ as upper bound of the sequence.

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Consider the Taylor series of $\sin(x)$:

$$\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n+1} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{3!}+\mathcal{O}(x^{6})$$

So:

$$\sin\left(\frac{1}{n^{2}+1}\right) = \frac{1}{n^{2}+1}-\mathcal{O}\left(\frac{1}{(n^{2}+1)^{3}}\right)$$

So Taylor expanding our limit:

$$\lim_{n\to\infty}n\sin\left(\frac{1}{n^{2}+1}\right) = \lim_{n\to\infty}\left(\frac{n}{n^{2}+1} - \mathcal{O}\left(\frac{n}{(n^{2}+1)^{3}}\right)\right) = 0$$

As you have found numerically.

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let $$k=\frac{1}{n}$$

So

$$\lim_{k \to 0}\frac{\sin(\frac{k^2}{1+k^2})}{k}$$ $$\lim_{k \to 0}\frac{k}{1+k^2}\frac{\sin(\frac{k^2}{1+k^2})}{\frac{k^2}{1+k^2}}=0$$

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