Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

The given function is $g(x)=x-[x]$.
It is evident that $g$ is defined at all integral points.
Let $n$ be an integer.
Then $g(n)=n-[n]=n-n=0$.
The left-hand limit of $g$ at $x=n$ is:
$\lim_{x \to n^-} g(x) = \lim_{x \to n^-} (x-[x]) = \lim_{x \to n^-} (x) - \lim_{x \to n^-} [x] = n - (n-1) = 1$.
The right-hand limit of $g$ at $x=n$ is:
$\lim_{x \to n^+} g(x) = \lim_{x \to n^+} (x-[x]) = \lim_{x \to n^+} (x) - \lim_{x \to n^+} [x] = n - n = 0$.
It is observed that the left-hand limit and right-hand limit of $g$ at $x=n$ do not coincide (since $1 \neq 0$).
Therefore,$g$ is not continuous at $x=n$.
Hence,$g$ is discontinuous at all integral points.