TITLE: $\lim \limits_{x \rightarrow 0 }\lfloor \cos(x) \rfloor= ?$
QUESTION [8 upvotes]: I was doing this question, i think the answer should be $0$ ,
My argument is as follows suppose $x \rightarrow 0$ from right side, that is $x $ approaches $0$ from positive side meaning $x$ is decreasing and therefore $\cos(x) $ should increase as $\cos(x)$ is a decreasing function , but $\cos(x) < 1$ hence their integral part will be $0$ , similarly for the negative case the $\cos(x)$ plays the same role as $\cos(x)$ is an even function. 
but at the same time i doubt what happens when $x = 0$,there $\cos(x) = 1$ and hence integer part will be $1$.
Please correct me if i am wrong.

REPLY [3 votes]: You're right.
The function $f(x)=\lfloor \cos x\rfloor$ is constant in a punctured neighborhood of $0$: for $x\in(-\pi/2,\pi/2)$, $x\ne0$,
$$
\lfloor \cos x\rfloor=0
$$
because $0<\cos x<1$. Therefore
$$
\lim_{x\to0}\lfloor \cos x\rfloor=0
$$
What the value of $\lfloor \cos x\rfloor$ is for $x=0$ is completely irrelevant as far as the limit is concerned, except for the fact you can conclude that $f$ is not continuous at $0$.