Determine if the floor function $f(x)=\lfloor x\rfloor$ is continuous at $1$.

Solution The floor function $f( x) =\lfloor x\rfloor =$ the greatest integer $\leq x$. The floor function $f$ is defined at $1$ and $f(1)=1$. But $$\begin{equation*} \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{-}}\lfloor x\rfloor =0\hbox{ and }\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1^{+}}\lfloor x\rfloor =1 \end{equation*}$$

$f(x)=\lfloor x\rfloor$

So, $\lim\limits_{x\rightarrow 1}\lfloor x\rfloor$ does not exist. Since $\lim\limits_{x\rightarrow 1}\lfloor x\rfloor$ does not exist, $f$ is discontinuous at $1$.