How it is possible that left side oscillation of function be 0 but left side limit does not exist? Let A be domain of function. The oscillation of function defined as: $$\Omega_{A} f:=\sup _{A} f-\inf _{A}=\sup _{x, y \in A}|f(x)-f(y)|$$
And oscillation of function at point defined as: $$\omega_{f}(c):=\lim _{h \rightarrow 0+} \Omega_{] c-h, c+h[\cap A} f$$
Show that if $$\lim _{x \rightarrow c-} f(x)$$ exsists  and is finite
then$$
\lim _{x \rightarrow c-}\omega_{f}(x)=0 $$
Show that converse is false.
Okey forward direction I understood. but I can't find any example that converse is false.
 A: For a counterexample of the converse, we want a function for which $\lim_{x\to c^-}\omega_f(x)=0$, but for which $\lim_{x\to c^-}f(x)$ either does not exist or is not finite.
I think the simplest counterexample would be any function which is continuous at all $x\ne c$, but for which $\lim_{x\to c^-}f(x)$ either does not exist or is not finite.
The first example that jumps to mind is $f(x)=1/x$ for $x\ne 0$ (and if we want $c=0$ in the domain, we can define $f(0)=0$)
To understand why this satisfies $\lim_{x\to c^-}\omega_f(x)=0$, we need to understand two things:
First, if $f$ is continuous at $x$, then $\omega_f(x)=0$. Therefore, for $f(x)=1/x$, we have that $\omega_f(x)=0$ for all $x \ne 0$ (here $c=0$).
Second, if any function satisfies $g(x) = a$ for $x<c$ [i.e. $g$ is constant on $(-\infty,c)$], then $\lim_{x\to c^-}g(x)=a$
Putting those two together, we have that $\lim_{x\to 0^-}\omega_f(x)=0$, but clearly $\lim_{x\to 0^-}f(x)$ is not finite.
For an example for which the limit does not exist, we could take
$$f(x) =
\begin{cases}
\sin(1/x) & x \ne 0 \\
0 & x=0
\end{cases}
$$
Again, we have that $f$ is continuous at $x\ne 0$, so that $\omega_f(x)=0$ for $x\ne 0$, and therefore that $\lim_{x\to 0^-}\omega_f(x)=0$. For this function $\omega_f(0)=2$.
