The function f(x) = (x), where (x) denotes the smallest integer ≥ x, is

continuous on R - Z

For any $n \in Z$, we have

(LHL at x = n) = $\lim\limits_{x \rightarrow n^{-}} f(x)$

⇒ (LHL at x = n) = $\lim\limits_{h \rightarrow 0} f(n-h)=\lim\limits_{h \rightarrow 0}(n-h)=n$

(RHL at x = n) = $\lim\limits_{x \rightarrow n^{+}} f(x)$

⇒ (RHL at x = n) = $\lim\limits_{h \rightarrow 0} f(n+h)=\lim\limits_{h \rightarrow 0}(n+h)=n+1$

∴  $\lim\limits_{x \rightarrow n^{-}} f(x) \neq \lim\limits_{x \rightarrow n^{+}} f(x)$

So, f(x) is discontinuous at x = n.

Thus, f(x) is not continuous at integer points.

Let $x=a \in R-Z$. Then, there exists $n \in Z$ such that $n<a<n+1$

Now,

$\lim\limits_{x \rightarrow a^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(a-h)=\lim\limits_{h \rightarrow 0}(a-h)=n+1$

$\lim\limits_{x \rightarrow a^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(a+h)=\lim\limits_{h \rightarrow 0}(a+h)=n+1$

and, $f(a)=n+1$

∴ $\lim\limits_{x \rightarrow a^{-}} f(x)=\lim\limits_{x \rightarrow a^{+}} f(x)=f(a)$

Thus, f(x) is continuous at x = a.

Hence, f(x) is continuous at all points other than integer points.