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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!

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    $\begingroup$ 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. $\endgroup$
    – user236182
    Nov 5, 2015 at 17:48
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    $\begingroup$ @user236182, although true, I don't think it's the main concern here. $\endgroup$
    – user175968
    Nov 5, 2015 at 17:49
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    $\begingroup$ @frank000 I never said it was the main concern, but it's faulty reasoning, so I'm pointing it out, so that they know. $\endgroup$
    – user236182
    Nov 5, 2015 at 17:50
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    $\begingroup$ The last one is correct. Recall as $x\to \infty$, $1/x \to 0$ and $\sin (1/x) \sim 1/x$. Thus, $x\sin(1/x)\sim x(1/x) =1$. $\endgroup$
    – Mark Viola
    Nov 5, 2015 at 17:51
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    $\begingroup$ Dr. MV is correct. Your friends are mistaken in saying that $\infty$*bounded = $\infty$. $0$ is bounded, and $\infty$ * $0$ is indeterminant. $\endgroup$
    – Paul Sinclair
    Nov 5, 2015 at 17:54

5 Answers 5

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

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  • $\begingroup$ So that means, that the second solution is correct, thanks! $\endgroup$
    – Atvin
    Nov 5, 2015 at 17:54
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    $\begingroup$ This answer is a copy of OP's second solution. $\endgroup$
    – user236182
    Nov 5, 2015 at 22:19
  • $\begingroup$ not $0$, need to be $0^+$. (Since limit exists, it is okay but not clear). $\endgroup$
    – student forever
    Jan 21, 2017 at 22:42
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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 $$\lim_{x \to \infty} f(x) g(x) = \lim_{x \to \infty} \frac{1}{x+1/x} = 0.$$

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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 $(1/y)\sin y\to1$.

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  • $\begingroup$ Thanks for your help! $\endgroup$
    – Atvin
    Nov 5, 2015 at 17:54
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    $\begingroup$ It's worth mentioning $\infty\times 0$ is an indeterminate form, which is why we can't determine the limit just by knowing it's of this form. $\endgroup$
    – user236182
    Nov 5, 2015 at 17:55
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Another way to see is

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

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  • $\begingroup$ This uses Taylor series, which requires knowing $\sin'(x)$. It's circular reasoning, because finding $\sin'(x)$ usually uses $\lim_{x\to 0}\frac{\sin x}{x}=1$. $\endgroup$
    – user236182
    Nov 5, 2015 at 21:03
  • $\begingroup$ @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. $\endgroup$
    – Federico Poloni
    Nov 6, 2015 at 1:05
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    $\begingroup$ @FedericoPoloni: that is exactly what you are calculating here, in a slightly disguised form $\endgroup$
    – Henry
    Nov 6, 2015 at 10:47
  • $\begingroup$ @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. $\endgroup$
    – Federico Poloni
    Nov 9, 2015 at 1:17
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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 $1$.

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  • $\begingroup$ Please explain why downvotes $\endgroup$
    – Kamil Jarosz
    Nov 6, 2015 at 12:42
  • $\begingroup$ I did not downvote, but people who read it quickly may read it as $0\cdot \infty=1$ and gave you a downvote. $\endgroup$
    – user175968
    Nov 6, 2015 at 21:22

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