why is limit $\lim_{x\rightarrow 0}⌊\frac{\sin x}{x}⌋ = 0$? I was evaluating the limit
$$f(x) = \lim_{x\rightarrow 0} \left\lfloor\frac{\sin x}{x}\right\rfloor$$
and I substituted the equivalent infinitesimal $\sin(x) \sim x$, obtaining $f(x) = 1$. But on observing the graph of $⌊\frac{\sin x}{x}⌋$, the limit comes out be $0$. However, evaluating $\lim_{x\rightarrow 0} ⌊\frac{\tan x}{x}⌋$ the same way, algebraic result matches with the graphical one. So what is going wrong in the first limit?
 A: Intuitively: Note that on $\lfloor \sin(x)/x \rfloor$, the graph approaches $1$ from below, while on $\lfloor \tan(x)/x \rfloor$, the graph approaches $1$ from above.
Therefore, while the graph of $\sin(x)/x$ approaches 1, it never reaches there: therefore, for small values of $x$ around 0, the value of $\lfloor \sin(x)/x \rfloor$ will be 0. However, even for small values of $x$ around 0, the value of $\lfloor \tan(x)/x \rfloor$ will be $1$, as it approaches from above and the values are greater than $1$. 
A: What is $$\lim_{x\to 0} \,\lfloor 1-x^2\rfloor?$$
It is not $1$. The problem is that $f(y)=\lfloor y\rfloor$ is not continuous.
A: For $x>0$ we have $\sin x<x$ (this can be verified by considering the derivative of $x-\sin x$). So, for $x>0$,
$$
\left\lfloor\frac{\sin x}{x}\right\rfloor=0
$$
Thus
$$
\lim_{x\to0^+}\left\lfloor\frac{\sin x}{x}\right\rfloor=0
$$
Since the function $\frac{\sin x}{x}$ is even, also
$$
\lim_{x\to0^-}\left\lfloor\frac{\sin x}{x}\right\rfloor=0
$$

For the other limit, the situation is different: for $0<x<\pi/2$ we have $\tan x>x$ (also this can be verified by using the derivative). Since
$$
\lim_{x\to0^+}\frac{\tan x}{x}=1
$$
there is $\delta>0$ such that, for $0<x<\delta$, $1<\frac{\tan x}{x}<2$ and so, for $0<x<\delta$,
$$
\left\lfloor\frac{\tan x}{x}\right\rfloor=1
$$
which gives, together with the fact that $\frac{\tan x}{x}$ is even,
$$
\lim_{x\to 0}\left\lfloor\frac{\tan x}{x}\right\rfloor=1
$$

In both cases you are actually doing the limit of a constant function (in a punctured neighborhood of $0$, for the second one).
