Find a function with left and right limits on [0,1] but with discontinuities

Hi. I am solving a question where I am asked to find a function with left and right limits on [0,1] but with discontinuities on A = {1/n : n in N}

I would be grateful if you can help me. I found the following function but I'm not sure I proved it has the right properties properly.

I got the function:


Then, I prove the right and left limits.

Let $x \in (0,1]$. Let $(x_m)$ be a sequence in $[0,1] \setminus \{x\}$ converging to $x$. Let $n$ be a natural number. Then there is a natural number $N$ such that for all $m > N$, we have $|x_m - x| < 1/n$.
Thus, for all $m > N$, $f(x_m) = x_m$. So $f(x_m)$ converges to $x$.

[The idea is that the sequence eventually gets so close to x that there are no points of the form 1/n in its neighbourhood. I'm not sure if I formalised it well.]

For $x = 0$, let $\epsilon > 0$. Let $(y_k)$ be a sequence in (0,1] converging to 0. Choose a natural number $K$ such that $y_k < \epsilon$ for all $k > K$. Then $f(y_m) \leq \epsilon$, so $f(y_m)$ converges to zero.

The function has discontinuities at 1/n, because by the above, for any c in (0,1] of the form 1/n, $lim_{x \rightarrow c}f(x) = c \neq 0 = f(c)$