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I was evaluating the limit $$f(x) = \lim_{x\rightarrow 0} \left\lfloor\frac{\sin x}{x}\right\rfloor$$ and I substituted the equivalent infinitesimal $\sin(x) \sim x$, obtaining $f(x) = 1$. But on observing the graph of $⌊\frac{\sin x}{x}⌋$, the limit comes out be $0$. However, evaluating $\lim_{x\rightarrow 0} ⌊\frac{\tan x}{x}⌋$ the same way, algebraic result matches with the graphical one. So what is going wrong in the first limit?

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    $\begingroup$ We have $\lvert \sin x \rvert < \lvert x\rvert$ for $x \neq 0$, but $\lvert x\rvert < \lvert \tan x \rvert$. $\endgroup$ – Daniel Fischer Jun 19 '15 at 20:35
  • $\begingroup$ what graph of sinx/x are you using ? $\endgroup$ – user247608 Jun 19 '15 at 20:51
  • $\begingroup$ Evaluating $\ldots$ the "sane" way or the "same" way? Is that a typo or a great coincidence? :-) $\endgroup$ – Asaf Karagila Jun 19 '15 at 21:11
  • $\begingroup$ Your argument would be valid if the function $u\mapsto\lfloor u\rfloor$ were continuous, but clearly it's not! $\endgroup$ – Michael Hardy Jun 19 '15 at 21:40
  • $\begingroup$ @AsafKaragila both apparently :) $\endgroup$ – Apoorv Jun 20 '15 at 9:59
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For $x>0$ we have $\sin x<x$ (this can be verified by considering the derivative of $x-\sin x$). So, for $x>0$, $$ \left\lfloor\frac{\sin x}{x}\right\rfloor=0 $$ Thus $$ \lim_{x\to0^+}\left\lfloor\frac{\sin x}{x}\right\rfloor=0 $$ Since the function $\frac{\sin x}{x}$ is even, also $$ \lim_{x\to0^-}\left\lfloor\frac{\sin x}{x}\right\rfloor=0 $$


For the other limit, the situation is different: for $0<x<\pi/2$ we have $\tan x>x$ (also this can be verified by using the derivative). Since $$ \lim_{x\to0^+}\frac{\tan x}{x}=1 $$ there is $\delta>0$ such that, for $0<x<\delta$, $1<\frac{\tan x}{x}<2$ and so, for $0<x<\delta$, $$ \left\lfloor\frac{\tan x}{x}\right\rfloor=1 $$ which gives, together with the fact that $\frac{\tan x}{x}$ is even, $$ \lim_{x\to 0}\left\lfloor\frac{\tan x}{x}\right\rfloor=1 $$


In both cases you are actually doing the limit of a constant function (in a punctured neighborhood of $0$, for the second one).

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  • $\begingroup$ this can't possibly be true, you can see that from using l'hopital's rule or expanding sinx (tanx ) by McLauren expansion, you see that sinx/x -> 1 $\endgroup$ – user247608 Jun 19 '15 at 20:49
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    $\begingroup$ @david Yes, I know that $\lim_{x\to0}\frac{\sin x}{x}=1$, but this has little to do with the problem at hand. The “floor” function is not continuous at $1$, so you can't “put the limit inside”. $\endgroup$ – egreg Jun 19 '15 at 20:51
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Intuitively: Note that on $\lfloor \sin(x)/x \rfloor$, the graph approaches $1$ from below, while on $\lfloor \tan(x)/x \rfloor$, the graph approaches $1$ from above.

Therefore, while the graph of $\sin(x)/x$ approaches 1, it never reaches there: therefore, for small values of $x$ around 0, the value of $\lfloor \sin(x)/x \rfloor$ will be 0. However, even for small values of $x$ around 0, the value of $\lfloor \tan(x)/x \rfloor$ will be $1$, as it approaches from above and the values are greater than $1$.

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What is $$\lim_{x\to 0} \,\lfloor 1-x^2\rfloor?$$

It is not $1$. The problem is that $f(y)=\lfloor y\rfloor$ is not continuous.

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