Disproving the continuity of a function I have the following question:

Suppose $f$ is a real-valued function on $\mathbb{R}$, and that, for a
given $x\in \mathbb{R}$, $\lim _{n \rightarrow
 \infty}\left[f\left(x+a_{n}\right)-f\left(x-a_{n}\right)\right]=0$, for all sequences $a_{n}$ that converge to $0$. Is $f$ continuous at $x$?

My answer is no. In support of this, I have constructed the following (putative) counter example:
$$f(x) \triangleq \begin{cases} 1 \hspace{5mm} x\neq0\\
0 \hspace{5mm}x=0\end{cases}$$
Note that, for $x=0$, $\lim_{n\rightarrow\infty}f(x+a_{n})=\lim_{n\rightarrow\infty}f(x-a_{n})$, for any $\{a_{n}\}_{n\in\mathbb{N}}$ which converges to $0$ (since $f(x)$ is an even function). Now, consider the sequence given by $a_{n}\triangleq \frac{1}{n}$. Since $f(a_{n})=1$ for all $n$, $\lim f(a_{n})=1.$ However, $f(0)=0$. Thus, $f$ is not continuous at $x=0$.
Thank you.

If we want to be absolutely strict - $\lim_{n \to \infty} f(x + a_n)$ doesn't have to exist - consider sequence $0, \frac{1}{2}, 0, \frac{1}{3}, 0, \frac{1}{4}, \ldots$ - or any other that contains infinite number of zero and infinite number of non-zero terms.
I don't think you need to even mention limits to prove that your $f$ satisfies this property: just note that $f(0 + a_n) - f(0 - a_n) \equiv 0$ and thus limit of it is $0$ too.
Your proof of $f$ not been continuous looks good.
