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The book I use is Jon Rogawski, multivariable calculus, chapter 1, question 39: evaluate lim {n (sin 1/n)}, for n→∞.

the student solution manual gives a fairly detailed explanation, it says: 1): lim (sinx)/x for x towards 0 = 1 (I understand that) 2): this implies lim (sin1/x)/(1/x) for x→∞, I do not understand this implication.

Also, when I try to graph this, it most certainly looks like a divergent function, so how could there possibly be a limit? I know that the value of sin 1/n only oscillates between -1 and +1, but the factor n keeps increasing.

Thanks for helping,

Roswitha

Roswitha
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2 Answers2

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Let's put $y=1/x$. As $x \rightarrow \infty$, $y \rightarrow 0$. So,

$\lim_{x \rightarrow \infty}{\sin(1/x)/(1/x)}=\lim_{y \rightarrow 0}\sin(y)/y$

voldemort
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Recalling Taylor series of $\sin(x)$ at $x=0$ we have

$$ n\sin(1/n) = n\left( \frac{1}{n}-\frac{1}{3!n^3}+\dots \right) .$$

Note: for large $n$ we have $\frac{1}{n}\sim 0$.

Mhenni Benghorbal
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    I usually do not like this argument, as even though individual terms in a series go to zero, the series itself might not be zero- there are no issues here, due to various reasons- but it might not be a good idea to tell a beginning calculus student about this, as it sometimes leads to wrong math. Sorry- this is my personal opinion :). – voldemort Aug 26 '14 at 03:53
  • @voldemort: It is a well known technique! – Mhenni Benghorbal Aug 26 '14 at 03:54
  • As I said- no issues here, but it's a matter of personal taste.. I generally advice caution; note that the caution is not for you- but for the beginning student. I am sure you know why this works- but sometimes students get into pitfalls while interchanging sums and limits. – voldemort Aug 26 '14 at 03:56