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I have researched the question $\lim_{n \to \infty} n*\sin(\frac{1}{n})$ quite profusely, and I know that it equals to 1, and I know why:

A) You can use a change of variables and substitute, say, $m = \frac{1}{n}$ so that $m \to 0$ instead.

B) L'Hopital's rule

The problem is, we haven't used either of these methods in class, so I am wondering if there is any other possible way to approach this question?

Inazuma
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  • There's always Taylor expansion – Lionel Ricci Apr 06 '16 at 22:42
  • Haven't really learnt that either :P – Inazuma Apr 06 '16 at 22:43
  • Have you used mean value theorem? – rtybase Apr 06 '16 at 22:54
  • @rtybase Pretty much all we've done is studied sequences and series (such as the ratio test, squeeze theorem etc.). The original question was actually to show whether the sum of $n\sin(\frac{1}{n})$ diverges (or converges), and my approach was to show that because lim $nsin(\frac{1}{n})$ doesn't equal 0, it must diverge. – Inazuma Apr 06 '16 at 22:59
  • See some ideas here http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – rtybase Apr 07 '16 at 18:08

1 Answers1

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To prove $\sin x/x,\,\tan x/x\to 1$ as $x\to 0$, consider the areas of a small-angle sector of a circle and the right-angled triangles obtained by using a radius for a hypotenuse or base. The squeeze theorem will complete the proof since $\cos x \to 1$.

J.G.
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