The $\frac{\sin x}x$ limit and the floor function $$\lim_{x\to0}\left\lfloor\frac{\sin x}x\right\rfloor=\,?$$
I think it should be 1, but in my class notes it's given as 0. I don't understand, because the limit without the floor brackets is 1, yet $\lfloor1\rfloor=1$.
 A: It is known that $|\sin x|\leq |x|,$ for $x \in \mathbb R$ and the equality holds only when $x=0$. Therefore for $x$ close to $0$, we have that:
$$0<\frac{\sin x}{x}<1\Rightarrow \left[\frac{\sin x}{x}\right]=0$$
and so the limit follows.
Note: Posted just a second after gobucksmath.
A: For $x\not=0$ it is always the case that $|x|>|\sin x|$, so we know that 
\begin{equation*}
\left|\frac{\sin x}{x}\right|<1 \text{ for }x\not=0
\end{equation*}
Applying the floor function will give you either $0$ or $-1$. For values of $x$ near $0$, the floor is $0$.
A: In a small neighbourhood around the origin, the function
$$
\frac{\sin x}{x}
$$
is less than $1$ and non-negative for all $x\ne 0$.  This means that for all $x\ne 0$ in that neighbourhood, we have
$$
\left\lfloor\frac{\sin x}{x}\right\rfloor=0
$$
and therefore the limit is $0$.  
As pointed out by Akiva Weinberger in the comments, the reason that your argument does not work is that limits are only necessarily preserved by continuous functions, and $\lfloor \cdot \rfloor$ is not continuous at $1$.  
Well done for being curious about this rather than just accepting what's in your class notes!
A: We have :
$$\forall x\in[-1,0[\,\cup\,]0,1],\quad 0\le\frac{\sin(x)}{x}<1$$
and therefore :
$$\forall x\in[-1,0[\,\cup\,]0,1],\quad \lfloor\frac{\sin(x)}{x}\rfloor=0$$
This implies that :
$$\lim_{x\to0}\lfloor\frac{\sin(x)}{x}\rfloor=0$$
A: Using the Taylor series of $\sin x$ around $x=0$ we get -
$$\frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac {x^4}{5!}-\frac{x^6}{7!} ....$$
Clearly
$$\frac{x^2}{3!}-\frac {x^4}{5!}+\frac{x^6}{7!} ....>0$$
since $|x^2|<1$ the numerators get smaller and smaller while the denominators keep on increasing(can be seen by taking terms in pairs). From this, we get
$$\frac{\sin x}{x} \rightarrow 1^{-} $$
as $x\rightarrow 0$ from either side. Therefore
$$\lim_{x \rightarrow0}\bigg\lfloor\frac{\sin x}{x}\bigg\rfloor = 0$$
