Bounded Piecewise Continuous Function may not have One-Sided Limits

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

Then: if $f$ satisfies Piecewise Continuous Function/Definition 1, $f$ satisfies Piecewise Continuous Function/Definition 2.

The reverse is not true.

Proof
Assume first that $f$ satisfies the requirements of Definition 1.

We need to prove that $f$ satisfies the requirements of Definition 2.

The only difference between definitions 1 and 2 lies in (2) in the two definitions, so we need to prove (2) in Definition 2, which requires that $f$ be bounded.

By Piecewise Continuous Function is Bounded, $f$ is indeed bounded, and this completes the proof that $f$ satisfies the requirements of Definition 2.

To prove that Definition 2 does not imply Definition 1, we need to find a function that satisfies Definition 2 but not Definition 1.

Consider the function


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

0 & x = a \\ \displaystyle sin \left( {\frac 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 $\displaystyle sin \left( {\frac 1 {x-a}} \right)$ is continuous on $\left({a \,.\,.\, b}\right)$, and since $f \left( x \right) = \displaystyle sin \left( {\frac 1 {x-a}} \right)$ on $\left({a \,.\,.\, b}\right)$ $f$ 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 $\displaystyle sin \left( {\frac 1 {x-a}} \right)$ varies between $-1$ and $+1$ as $x$ approaches $a$ from above and so 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, and so requirement (2) of Definition 1 is not satisfied.