Bounded Piecewise Continuous Function may not have One-Sided Limits

Theorem
Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.

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:


 * $\map f x = \begin{cases}

0 & : x = a \\ \map \sin {\dfrac 1 {x - a} } & : x \in \hointr a b \end{cases}$

Consider the (finite) subdivision $\set {a, b}$ of $\closedint a b$.

We observe that $\map \sin {\dfrac 1 {x - a} }$ is continuous on $\openint a b$.

Since $\map f x = \map \sin {\dfrac 1 {x - a} }$ on $\openint a b$, it follows that $f$ is continuous on $\openint a b$.

Also, $f$ is bounded by the bound $1$ on $\closedint a b$.

Therefore $f$ is a bounded piecewise continuous function on the closed interval $\closedint a b$.

We now investigate whether $f$ has one-sided limits at the endpoints of $\closedint a b$.

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

Thus it does not converge.

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

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

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