TITLE: why is limit $\lim_{x\rightarrow 0}⌊\frac{\sin x}{x}⌋ = 0$?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.