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What is the value of $\Bigl\lfloor\,\lim_{x\to 0}\frac{\sin x}{x}\Bigr\rfloor$? Is it $1$ or $0$?

I was told that the answer is $0$ as $\sin{x}$ is less than $x$ as $x\rightarrow0$. Is it correct or are limits exact values?

I know that $\lim_{x\to0}\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor$ will be $0$ due to the above mentioned fact.

user
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Harshit Joshi
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    $$ \lfloor 1 \rfloor = 1 \quad\text{and}\quad \lim_{x\to 0}\frac{\sin x}{x} = 1 $$ – MSDG Sep 04 '18 at 11:51
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 04 '18 at 11:58
  • @Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting. – Harshit Joshi Sep 04 '18 at 12:01

3 Answers3

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We have $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{\cos x}{1}=\cos (0) = 1.$$ Since $\lfloor 1\rfloor=1$ it follows that $\left\lfloor\left(\lim_{x\to 0}\left(\frac{\sin x}{x}\right)\right)\right\rfloor=1$.

Arnaud D.
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YukiJ
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Since

  • $\lim_{x\to0}(\frac{\sin x}{x})=1 $
  • and for $x\ne 0$ sufficiently small: $0<\frac{\sin x}{x}<1$

we have that

$$\Bigl\lfloor\,\lim_{x\to0}\frac{\sin x}{x}\Bigr\rfloor=1$$

and

$$\lim_{x\to0}\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor=0$$

user
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  • Thanks, but the question is answered already. – Harshit Joshi Sep 04 '18 at 13:11
  • @HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases. – user Sep 04 '18 at 13:15
  • Why is it wrong? – Harshit Joshi Sep 04 '18 at 13:17
  • @HarshitJoshi As you mentioned for any $x\ne 0$ we have $0<\frac{\sin x}{x}<1$ and therefore $$\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor=0 \implies \lim_{x->0}\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor=0$$ – user Sep 04 '18 at 13:20
  • Still don't understand why is it wrong? – Harshit Joshi Sep 04 '18 at 13:22
  • He is talking about the left and right hand limits. – Harshit Joshi Sep 04 '18 at 13:23
  • @HarshitJoshi Note that $$\Bigl\lfloor,\lim_{x->0}\frac{\sin x}{x}\Bigr\rfloor=\Bigl\lfloor,\lim_{x->0^+}\frac{\sin x}{x}\Bigr\rfloor=\Bigl\lfloor,\lim_{x->0^-}\frac{\sin x}{x}\Bigr\rfloor=1$$ in any case the limit is exactly 1 and therefore the floor of the limit is 1 (but the limit of the floor is 0). – user Sep 04 '18 at 13:27
  • Yes you are right. Did not look carefully. – Harshit Joshi Sep 04 '18 at 13:30
  • @gimusi I think even the floor of the limit is 0 , because $\lim_{x\to 0}\frac{\sin x}{x}= 0.999999\ldots $, but it is never equal to 1 .We cannot define the function when denominator is 0. We define only the $\lim_{x\to 0}$ – Awakened Sep 04 '18 at 16:06
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By the Maclaurin's expansion of $sin(x)$, we have, $$\sin(x) =\sum^{\infty}_{k=0}\frac{(-1)^k}{(2k+1)!}x^{2k+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

As $x\to0$ we compute left hand limit, $$\lim_{x\to0^{-}}\frac{\sin(x)}{x} =\lim_{x\to0^{-}}\left( 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right) $$ $$\lim_{x\to0^{-}}\frac{\sin(x)}{x} = 1$$ Further for right hand limit, $$\lim_{x\to0^{+}}\frac{\sin(x)}{x} =\lim_{x\to0^{+}}\left( 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right) $$ $$\lim_{x\to0^{+}}\frac{\sin(x)}{x} = 1$$ Now, $$\Bigl\lfloor\lim_{x\to0^{+}}\frac{\sin(x)}{x}\Bigr\rfloor = \Bigl\lfloor\lim_{x\to0^{-}}\frac{\sin(x)}{x}\Bigr\rfloor =1 $$

Sahil Silare
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