What is the limit of $n\sin \left(\frac{1}{n}\right)$ as $n \to \infty$

Question

The textbook I'm reading gives no proof of the limit being 1 apart from a flimsy 0/0, and has covered only the "epsilon" and sandwich test for convergence to a particular limit.

Do not rearrange to turn $n$ into $\frac{1}{n}$, that is the source of the flimsy $0/0$ argument.

See How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? — Martin Sleziak, Dec 09 '14 at 17:17

score 4 · Accepted Answer · answered Dec 09 '14 at 08:42

4

Hint: $$\lim_{x\to 0}\frac{\sin x}x=1.$$

answered Dec 09 '14 at 08:42

Martín-Blas Pérez Pinilla

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But $why$? All I can see is 0/0...sandwich rule? Epsilons? – Nethesis Dec 09 '14 at 08:48
@You can find the explanation why this is true in just about any textbook on calculus... – 5xum Dec 09 '14 at 08:53
There is nothing so far mentioned that implies that is anything more than naive intuition – Nethesis Dec 09 '14 at 08:53
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@Nethesis Here, let me google that for you: http://www.ies-math.com/math/java/calc/LimSinX/LimSinX.html – 5xum Dec 09 '14 at 08:57
Thank you, that's what I was looking for – Nethesis Dec 09 '14 at 09:11

DeepSea · Answer 2 · 2014-12-09T09:44:37.970

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Hint: $\sin \left(\dfrac{1}{n}\right) \approx_{0} \dfrac{1}{n} - \dfrac{1}{2!}\cdot \dfrac{1}{n^2} + \dfrac{1}{3!}\cdot \dfrac{1}{n^3}- ......\Rightarrow n\sin \left(\frac{1}{n}\right) \approx_0 1 - \dfrac{1}{2n} + \dfrac{1}{6n^2} - ....--$

edited Dec 09 '14 at 09:44

answered Dec 09 '14 at 08:33

DeepSea

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That's...a lot more satisfactory – Nethesis Dec 09 '14 at 08:49
@Nethesis A warning for this answer: it uses more advanced mathematics than L'Hospital's rules. In effect, if you are using this, you are using a mathematical tool (the Taylor series) from which L'Hospital trivially follows, so if you don't yet know how to use L'Hospital, don't use this. – 5xum Dec 09 '14 at 08:55
I know, that's why I'm not completely satisfied – Nethesis Dec 09 '14 at 09:04

What is the limit of $n\sin \left(\frac{1}{n}\right)$ as $n \to \infty$

2 Answers2