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$$\lim_{x\to0}\left\lfloor\frac{\sin x}x\right\rfloor=\,?$$ 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 $\lfloor1\rfloor=1$.

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    The floor function is not continuous, so it's not true that $\lim\lfloor f(x)\rfloor=\lfloor\lim f(x)\rfloor$. – 2017-01-13
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    For all $0<|x|<\pi$, the function is $0$. – 2017-01-13
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    To see what's happening, and understand the answers below, it might help to draw the circle $x^2 + y^2 =1$ (please note I've changed the meaning of $x$ from that in your question/ other answers): for $\theta>0$, $\sin \theta /\theta $ is the ratio of the $y$ value over the length of the arc on the circle subtended by $\theta$. Pictorially, the arc is longer than $y$, so the (positive) ratio is (always) less than $1$, so the floor is always $0$. Therefore the limit must also be $0$. – 2017-01-13

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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.

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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 $0$.

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    Thanks ! But we were thought that Lt x-->0 (sin x/x) = 1. If x is tending to zero it means it's not equal to zero, so shouldn't x-->0 (sin x/x) < 1 be the correct one? – 2017-01-13
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    Answer this question: what is $\lim_{x\to 0}\lfloor 1-|x|\rfloor$? – 2017-01-13
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    Left hand limit - Lim x--> 0 [ 1 + x ] ===> [ 1+ 0 ] = 1, Right hand limit - Lim x --> 0 [ 1- x ] ====> [ 1-0 ] = -1. So limit does not exist. – 2017-01-13
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    When evaluating limits, you only care about values of your function close to the limit value. Take $x=0.1, 0.01, 0.001...$. What happens? – 2017-01-13
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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!

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    Not necessarily true for all $x \neq 0$, since once you get far enough from the origin, you'll get negative values. Better to say that it's true in the neighborhood of the origin. – 2017-01-13
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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 :

$$\lim_{x\to0}\lfloor\frac{\sin(x)}{x}\rfloor=0$$