Why does $\lim_{x \to 0} \lfloor n \cdot \frac{\sin x}{x} \rfloor = n-1$? I am given the following solution but I don't understand why does the greatest integer value of a natural number multiplied to a value which is less than $1$ is coming out to be $n-1$. Can you explain why?

Given $n \in \mathbb{N}$, evaluate $\lim_{x \to 0} \left( \lfloor n \cdot \frac{\sin x}{x}\rfloor + \lfloor n \cdot \frac{\tan x}{x} \rfloor \right)$

Solution:

Since $\sin x < x$ and $\tan x > x$, the quantity of interest is equal to $(n-1) + n = 2n-1$.

 A: Note that we also have $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\tan x}{x}=1$. They are arbitrarily close to $1$ at the neighborhood of $x=0$. Also, around that neighborhood besides $x=0$, we have $\frac{\sin x}{x}$ being strictly less than $1$ but it can be made arbitarily close to $1$ and $\frac{\tan x}{x}$ being strict greater than $1$ but it can get arbitrarily close to $1$.
Hence $\lim_{x \to 0} \lfloor n \cdot \frac{\sin x}{x}\rfloor = n-1$ since I can make $x$ close enough that $n-1 < n \cdot \frac{\sin x}{x}< n$. I just have to consider a neighborhood near $0$, $x \ne 0$, $1-\frac1n < \frac{\sin x}{x} < 1$.
and $\lim_{x \to 0} \lfloor n \cdot \frac{\tan x}{x}\rfloor = n$ since I can make $x$ close enough that $n < n \cdot \frac{\tan x}{x} < n+1$. I just have to consider a neighborhood near $0$, $x \ne 0$, $1 < \frac{\tan x}{x} < 1+\frac1n$.

A: For $x\ne 0$, $$\frac{\sin x}x<1\text{ and }\frac{\tan x}x>1$$  so that $$\left\lfloor n\frac{\sin x}x\right\rfloor<n\text{ and }\left\lfloor n\frac{\tan x}x\right\rfloor\ge n.$$
On another hand, for $x$ sufficiently close to $0$,
$$\left\lfloor n\frac{\sin x}x\right\rfloor\ge n-1\text{ and }\left\lfloor n\frac{\tan x}x\right\rfloor<n+1.$$

In other words, for tiny $\epsilon$ (smaller than $\frac1n$ but positive),
$$\lfloor n(1-\epsilon)\rfloor=n-1\text{ and }\lfloor n(1+\epsilon)\rfloor=n.$$
