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I need to find a limit of a sequence: $$\lim_{n\to\infty}n\frac{\sin\frac{1}{n}-\frac{1}{n}}{1+\frac{1}{n}}$$

I tried to divide numerator and denominator by n, but it didn't help, as the limit became $\frac{0}{0}$. I tried other things, but always got an indefinite limit. I know that the limit is 0, but I just don't know how to show it. It's probably something really simple, but I'm totally stuck.

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    $\begingroup$ Hint: it doesn't go to 0. $\endgroup$ – user80944 Dec 2 '13 at 21:37
  • $\begingroup$ Oh, I'm sorry, I put it in wrong! I'll edit it! $\endgroup$ – Robert Dec 2 '13 at 21:38
  • $\begingroup$ Is this not -1? $\endgroup$ – Alec Teal Dec 2 '13 at 21:39
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    $\begingroup$ The bottom seems irrelevant, it may still be typed incorrectly. I would use Maclaurin series, though L'Hospital's Rule also works well. $\endgroup$ – André Nicolas Dec 2 '13 at 21:42
  • $\begingroup$ The whole numerator, sorry if I put it in wrong $\endgroup$ – Robert Dec 2 '13 at 21:42
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HINT

Do you know about $\lim\limits_{x\to 0}\frac 1 x \sin (x) $ ?

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  • $\begingroup$ I don't think I do, I'm trying to figure out Daniel's answer right now. $\endgroup$ – Robert Dec 2 '13 at 21:52
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    $\begingroup$ @Robert : got to know that, it is the microwave solution to many limit problems. $\endgroup$ – Arjang Dec 3 '13 at 0:10
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We have $$\lim_{n\to\infty}n\frac{\sin\frac{1}{n}-\frac{1}{n}}{1+\frac{1}{n}} = \lim_{x\to 0}\frac{\frac{\sin x}{x} - 1}{1+x}$$ Using the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$, we get the result $$\lim_{x\to 0}\frac{\frac{\sin x}{x} - 1}{1+x}=\frac{1-1}{1}=0$$

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