Bounded Piecewise Continuous Function may not have One-Sided Limits

Theorem
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$, $a < b$.

Let $f$ be a bounded piecewise continuous function.

Then it is not necessarily the case that $f$ is a piecewise continuous function with one-sided limits:

Proof
Consider the function:


 * $f \left({x}\right) = \begin{cases}

0 & : x = a \\ \sin \left({\dfrac 1 {x - a} }\right) & : x \in \left({a \,.\,.\, b}\right] \end{cases}$

Consider the (finite) subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.

We observe that $\sin \left({\dfrac 1 {x - a} }\right)$ is continuous on $\left({a \,.\,.\, b}\right)$.

Since $f \left({x}\right) = \sin \left({\dfrac 1 {x - a} }\right)$ on $\left({a \,.\,.\, b}\right)$, it follows that $f$ is continuous on $\left({a \,.\,.\, b}\right)$.

Also, $f$ is bounded by the bound $1$ on $\left[{a \,.\,.\, b}\right]$.

Therefore $f$ is a bounded piecewise continuous function on the closed interval $\left[{a \,.\,.\, b}\right]$.

We now investigate whether $f$ has one-sided limits at the endpoints of $\left[{a \,.\,.\, b}\right]$.

The function $\sin \left({\dfrac 1 {x - a} }\right)$ varies between $-1$ and $+1$ as $x$ approaches $a$ from above.

Thus it does not converge.

Since $f \left({x}\right) = \sin \left({\dfrac 1 {x - a} }\right)$ when $x > a$ we conclude that $\displaystyle \lim_{x \mathop \to a^+} f \left({x}\right)$ does not exist either.

$f$ has no one-sided limit at $a$.

Hence $f$ is not a piecewise continuous function with one-sided limits.