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$$\lim_{x\to\infty} x \sin\left(\frac{1}{x}\right) = ?$$

Not a long ago I saw this function, and I was curious, what limit it has, when $x$ approaches $\infty$? Some of my friends said fast that it must approach $\infty$, since $\sin$ is a bounded function, and $x$ goes to infinity, therefore infinity * bounded must be infinity. Some others said that $\sin(\frac{1}{x})$ is $0$, since $\frac{1}{x}$ is $0$, when $x \rightarrow \infty$.

So, the first possible solution should be $\infty$, but here is an other one. Let $y=\frac{1}{x}$. If $x \rightarrow \infty$, then $y \rightarrow 0$. Using that:

$$\lim_{x\to\infty} x \sin\left(\frac{1}{x}\right) = \lim_{y\to 0} \frac {1}{y} \sin(y) = \lim_{y\to 0} \frac{\sin(y)}{y} = 1$$

Here is a proof that, $\lim_{y\to 0} \frac{\sin(y)}{y} = 1$: Proof

So here we have $2$ completely different solutions for the same task, which both seem "logical". Is any of them correct, or if not, what should be the solution? Is this convergent, or divergent? Any help appreciated!

Community
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Atvin
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    Using L'Hospital's Rule on $\lim_{y\to 0}\frac{\sin y}{y}$ is circular reasoning, because the derivation of $\sin'(x)$ usually uses that limit. – user236182 Nov 05 '15 at 17:48
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    @user236182, although true, I don't think it's the main concern here. –  Nov 05 '15 at 17:49
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    @frank000 I never said it was the main concern, but it's faulty reasoning, so I'm pointing it out, so that they know. – user236182 Nov 05 '15 at 17:50
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    The last one is correct. Recall as $x\to \infty$, $1/x \to 0$ and $\sin (1/x) \sim 1/x$. Thus, $x\sin(1/x)\sim x(1/x) =1$. – Mark Viola Nov 05 '15 at 17:51
  • Why infinity*bounded must be infinity? $\lim_{x\to \infty} x\cdot 0=0$. –  Nov 05 '15 at 17:53
  • I edited the question with the limit rule I used, but @frank000 is right, that is not the main question, thanks for pointing it out though. – Atvin Nov 05 '15 at 17:53
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    Dr. MV is correct. Your friends are mistaken in saying that $\infty$bounded = $\infty$. $0$ is bounded, and $\infty$ $0$ is indeterminant. – Paul Sinclair Nov 05 '15 at 17:54

5 Answers5

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$$\lim_{x\to \infty }x\sin \frac{1}{x}=\lim_{x\to\infty }\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}=\lim_{u\to 0}\frac{\sin(u)}{u}=1$$

Surb
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The first line of reasoning is clearly incorrect: let $f(x) = x$, $g(x) = (x^2 + 1)^{-1}$. Then $$\lim_{x \to \infty} f(x) = \infty$$ and $g(x) \in (0,1]$ is clearly bounded on $\mathbb R$. But $$\lim_{x \to \infty} f(x) g(x) = \lim_{x \to \infty} \frac{1}{x+1/x} = 0.$$

heropup
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$\infty\times\text{bounded}$ is not always $\infty$. You provide an example in your question. As $y\to0$, $1/y\to\infty$ and $\sin y$ is bounded. But $(1/y)\sin y\to1$.

Julián Aguirre
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Another way to see is

$$\lim_{x\to \infty} x\sin{\frac{1}{x}}=\lim_{x\to \infty} x\{\frac{1}{x}-\frac{1}{3!x^3}+\frac{1}{5!x^5}-\cdots\}=\lim_{x\to \infty}\{1-\frac{1}{3!x^2}+\frac{1}{5!x^4}-\cdots\}=1$$

Kushal Bhuyan
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  • This uses Taylor series, which requires knowing $\sin'(x)$. It's circular reasoning, because finding $\sin'(x)$ usually uses $\lim_{x\to 0}\frac{\sin x}{x}=1$. – user236182 Nov 05 '15 at 21:03
  • @user236182 Why can't we take knowledge of $\sin' x $ for granted? We are not computing $\lim_{x\to 0} \frac{\sin x}{x}$ here, but solving a different problem. – Federico Poloni Nov 06 '15 at 01:05
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    @FedericoPoloni: that is exactly what you are calculating here, in a slightly disguised form – Henry Nov 06 '15 at 10:47
  • @Henry No, it is not the same problem. This is an exercise. It is crucial that one does not use circular arguments while developing and explaining the theory. But exercises and problems can take for granted the full theory explained in a textbook, since it is already existing and proved in a self-consistent way (unless this is one of those contrived "compute this limit without using..." exercises, in which case OP should have specified it). All proofs in this thread use the known result $\lim_{x\to 0} \frac{\sin x}{x}=1$, and there is no reason to avoid it in an exercise. – Federico Poloni Nov 09 '15 at 01:17
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You faced here an indeterminate form. It means that

$$0\cdot\infty$$

may be equal to $0$, $\infty$, any other number or may not exist at all. In this case it equals $1$.

Kamil Jarosz
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