Discontinuity - Unsure If Piecewise Equation(s) Have Them I have a question on whether the functions following have a discontinuity, and if not, what are the points where two functions meet. 
First, the piece wise equation :
\begin{align*}
f(x)=
\begin{cases}
\sin(x), &\text{ if } 0 \leq x \leq 2\pi;\\
0, &\text{ if } x<0 \text{ or } x>2\pi.
\end{cases}
\end{align*}
What are the points called at $0$ and $2\pi$ ?. Are they a discontinuity, as I expect that they are not since there is no jump. 
The second piece wise equation is :
\begin{align*}
f(x)=
\begin{cases}
\sin(x), &\text{ if }0 \leq x \leq \pi/2;\\
0, &\text{ if }  x<0;\\
1, &\text{ if }      x>\pi/2.
\end{cases}
\end{align*}
What is the point called at $\pi/2$? Is it a discontinuity, and if not, how would they be described ?
The reason for this question is that I am examining DSP filters – and I need to ensure that I use the correct terminology when documenting results, and also, to ensure that I understand the theory. There have been statements that the first equation does have discontinuities, but I examined the definition, and I am unsure. 
Thanks and regards,
Code_X. 
 A: I don't know if the points where the components of piece wise functions join have a name, but checking whether they are discontinuities, or non-differentiable points is not difficult.
Lets begin with discontinuities. In order for the point $p$ not to be a discontinuity, $$\lim_{x\to p^+} f(x) = \lim_{x\to p^-} f(x) = f(p)$$
In the case of $p=0$, for your first piece-wise function, since $\sin(x)$ is continuous, $$f(0) = \lim_{x\to0^+} f(x)$$
So all that needs to be shown is that $\lim_{x\to0^-} f(x)=f(0)$, which is obvious since $$\forall x<0, f(x)=0$$
For the second piecewise function, the exact same logic applies since the functions at this point are the same.
You can apply similar logic to the second point joining point of the piece-wise function, but of course, the cases of both functions are a tad bit different there.  See how far you can get!
On the other hand, if we find a point to be continuous (i.e., $f(0)$), we can also check whether we can find the derivative of the point, or if the point is a non-differentiable point on the function.
We do that with the following formula :$$f'(p)=\lim_{h\to 0}\frac{f(p+h)-f(p)}{h}$$ If this exists, then it is differentiable. This exists iff $$\lim_{h\to 0^+}\frac{f(p+h)-f(p)}{h}=\lim_{h\to 0^-}\frac{f(p+h)-f(p)}{h}$$
In the case of $p=0$, when $h>0$, $f(p+h)=sin(h)$ and $f(p)=sin(p)=0$, so  $$\lim_{h\to 0^+}\frac{f(p+h)-f(p)}{h} = \lim_{h\to 0^+}\frac{\sin(h)}h = 1$$
However, when $h<0$, $f(p+h)= 0$ and $f(p) = 0$, so $$\lim_{h\to 0^-}\frac{f(p+h)-f(p)}{h} = \frac{0-0}h = 0$$ Since these one sided limits are not equal, the function is not differentiable at $p=0$. Again, you can use this logic on the other point where the piece-wise functions join.
