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 $f$ satisfy Definition 1 of a piecewise continuous function:

Then $f$ satisfies Definition 2 of a piecewise continuous function.

The converse is not true.

Proof
Let $f$ satisfy the requirements of Definition 1 of a piecewise continuous function.

It is to be proved that $f$ satisfies the requirements of Definition 2 of a piecewise continuous function.

The only difference between definition 1 and definition 2 lies in $(2)$ in the two definitions.

Thus we need to prove $(2)$ in definition 2.

This requires that $f$ be bounded.

By Piecewise Continuous Function is Bounded, $f$ is indeed bounded.

This completes the proof that $f$ satisfies the requirements of Definition 2.

It remains to be proved that definition 2 does not necessarily imply definition 1.

Our strategy is to find a function that satisfies definition 2 but not definition 1.

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 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 $f$ satisfies the requirements of Definition 2 for the subdivision $\left\{{a, b}\right\}$.

We now consider $(2)$ of Definition 1, which requires in particular that $\displaystyle \lim_{x \to a+} f \left({x}\right)$ exist.

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 $\displaystyle \lim_{x \to a+} f\left({x}\right)$ does not exist either.

So requirement $(2)$ of Definition 1 is not satisfied.