$\lim_{x\to 0}\left\lfloor\frac{n\sin(x)} x \right\rfloor,n\in\mathbb N$ If $\lfloor\,\cdot\,\rfloor$ denotes the greatest integer function $n\in\mathbb N$, then what is the value of 
$$\lim_{x\to 0}\left\lfloor\frac{n\sin(x)} x \right\rfloor \text{ ?}$$
My try:
\begin{align*} \lim_{x\to 0} \left\lfloor\frac{n\sin(x)} x \right\rfloor &= \lim_{x\to 0} \left\lfloor \frac{n(x-x^3/3!+x^5/5!-\cdots)} x \right\rfloor \\[10pt]
&= \lim_{x\to 0} \left\lfloor \frac{n(1-x^2/3!+x^4/5!-\cdots)} 1 \right\rfloor \\[10pt]
&= \lfloor n\rfloor \end{align*}
But the answer is $n-1$.
How am I committing a mistake? Why can't limits enter into the greatest integer function? What is the correct solution for this problem?
 A: Consider that $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$ and that $\frac{\sin(x)}{x} < 1$ for all $x\in \mathbb{R}\setminus \{0\}$. Therefore, for any $\epsilon > 0$, there is some $\delta > 0$ such that $0 < \lvert x\rvert < \delta$ implies $1-\epsilon < \frac{\sin(x)}{x} < 1$. From this, we can see that for any $\epsilon > 0$, we can choose the same $\delta > 0$ to find that $0 < \lvert x\rvert < \delta$ implies $$(1-\epsilon)n < \frac{n\sin(x)}{x} < n$$ Namely, this holds true for $\epsilon < \frac{1}{n}$ (which gives $(1-\epsilon)n > n-1$), so the result follows, i.e. $$\lim_{x\to 0} \left\lfloor \frac{n\sin(x)}{x}\right\rfloor = n-1$$
A: For $x\ne0$, $$\dfrac{\sin x}x<1$$ so that $$\dfrac{n\sin x}x<n.$$
For all $x$ sufficiently close to $0$, the floor is $n-1$* (but never $n$), and this is the requested limit.

*By continuity of the function you will always find an interval where this holds.
A: Maybe lots of students recall that $\dfrac{\sin x} x \to 1$ as $x\to0,$ but if you recall the proof of that fact you see that $\dfrac{\sin x} x \uparrow 1$ as $x\to0,$ i.e. it approaches $1$ from below. Therefore $\dfrac{n\sin x} x \uparrow n,$ i.e. this approaches $n\vphantom{\frac {\sum^\sum}{}}$ from below.
