I have recently stumbled upon the following function $f [0,1] \rightarrow \mathbb{R}$:
$$ f(x) = \begin{cases} e, & \text{if } x = 0 \\ \sum_{n=0}^y \frac{1}{n!}, & \text{if } x>0 \end{cases} $$ where $y = [\frac{1}{x}]$ is the integer part of $\frac{1}{x}$. The claim is that it is Riemann integrable but has infinitely many discontinuities. I clearly can see that it is Riemann Integrable since it is a decreasing function on the given interval $[0,1]$. However, I am not sure about the proof of being infinitely discontinuous. Here is the proof that came with this example:
Take $n \in \mathbb{N}$ and set $x_0 = \frac{1}{n}$. Then, $\lim_{x\to x_0^-} f(x) - \lim_{x\to x_0^+} f(x) = \frac{1}{n!}$
Here I understand that discontinuity follows, however, I am not sure how they obtained that the difference is $\frac{1}{n!}$.
Thank you in advance!
We are seeing the step discontinuity.
Let $x_{0}=\frac{1}{n}=\frac{1}{3}$. As we approach from the left, $$\color{blue}{2.9,\ 2.99,\ 2.999},\dots$$ The function value is $$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}$$ As we approach from the right, $$\color{red}{ 3.1,\ 3.01,\ 3.001}$$ The function value is $$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \boxed{\frac{1}{3!}}$$
The difference is $\frac{1}{n!} = \frac{1}{3!}$