The floor function is discussed in Section P.2, p. 17.

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 \(1\).