-1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ $$ \lfloor 1 \rfloor = 1 \quad\text{and}\quad \lim_{x\to 0}\frac{\sin x}{x} = 1 $$ $\endgroup$ – MisterRiemann Sep 4 '18 at 11:51
  • 1
    $\begingroup$ 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
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.