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.

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    $\begingroup$ $$ \lfloor 1 \rfloor = 1 \quad\text{and}\quad \lim_{x\to 0}\frac{\sin x}{x} = 1 $$ $\endgroup$ – MisterRiemann Sep 4 '18 at 11:51
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    $\begingroup$ 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. $\endgroup$ – Shaun Sep 4 '18 at 11:58
  • $\begingroup$ @Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting. $\endgroup$ – harshit54 Sep 4 '18 at 12:01

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 $\left\lfloor\left(\lim_{x\to 0}\left(\frac{\sin x}{x}\right)\right)\right\rfloor=1$.

  • $\begingroup$ So limits are exact values and not values tending to a number, am i right? $\endgroup$ – harshit54 Sep 4 '18 at 11:59
  • $\begingroup$ @nicomezi So are limits not exact values? $\endgroup$ – harshit54 Sep 4 '18 at 12:02
  • $\begingroup$ If they exist, yes. However, it might be that a limit does not exist (for example $\lim_{x \to \infty} x$ does not exist). $\endgroup$ – YukiJ Sep 4 '18 at 12:02
  • $\begingroup$ I missread the question. The answer is correct. $\endgroup$ – nicomezi Sep 4 '18 at 12:03
  • $\begingroup$ use \to instead of -> for tending $\endgroup$ – Deepesh Meena Sep 4 '18 at 12:30


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

we have that

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


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

  • $\begingroup$ Thanks, but the question is answered already. $\endgroup$ – harshit54 Sep 4 '18 at 13:11
  • $\begingroup$ @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. $\endgroup$ – user Sep 4 '18 at 13:15
  • $\begingroup$ Why is it wrong? $\endgroup$ – harshit54 Sep 4 '18 at 13:17
  • $\begingroup$ @HarshitJoshi As you mentioned for any $x\ne 0$ we have $0<\frac{\sin x}{x}<1$ and therefore $$\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor=0 \implies \lim_{x->0}\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor=0$$ $\endgroup$ – user Sep 4 '18 at 13:20
  • $\begingroup$ Still don't understand why is it wrong? $\endgroup$ – harshit54 Sep 4 '18 at 13:22

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, $$\Bigl\lfloor\lim_{x\to0^{+}}\frac{\sin(x)}{x}\Bigr\rfloor = \Bigl\lfloor\lim_{x\to0^{-}}\frac{\sin(x)}{x}\Bigr\rfloor =1 $$


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