Find two functions $f$ and $g$ such that they are both discontinuous at $c$, however, $f+g$ and $f\cdot g$ are both continuous at $c$ Could someone please explain to me how to approach these kinds of question and also what is the answer to the following question?
Give an example of a function $f$ and $g$ such that they are both discontinuous at $c$, however, $f+g$ and $f\cdot g$ are both continuous at $c$.
 A: When given such questions, it is easy to think of piecewise defined functions that by themselves have discontinuities (most commonly the discontinuity is at the point where the piecewise definition "splits"), but their sum / product ends up being constant, and thus continuous.
For example, let
$$
f(x) = \begin{cases}
1 & \text{if }x \ge 0 \\
-1 & \text{if } x < 0
\end{cases}
$$
and
$$
g(x) = \begin{cases}
-1 & \text{if }x \ge 0 \\
1 & \text{if } x < 0
\end{cases}
$$
In other words, $f(x) = \operatorname{sgn}(x)$ and $g(x) = -\operatorname{sgn}(x)$. It is easy to see that $(f + g)(x) = 0 \; (\forall x \in \mathbb{R})$, and so is continuous at $0$, too.
Can you think of a similar example for your second question? Think of two functions that are continuous everywhere except at $0$, but their product evaluates to a constant everywhere. Or you can just look at $f$ and $g$ as I defined them previously :)
A: Let $f(x)$ be 0 everywhere but at $x_1$, where $f(x)=1$ 
Let $g(x)= 1$ everywhere but at $x_1$ where $g(x)= 0$ 
Then $f+g$= 1 everywhere and $f*g$= 0 everywhere 
A: One of the simplest functions that is discontinuous at a point is a step function. Can you see how to get the discontinuities to cancel in the sum? For the product if one or the other is $0$ everywhere, what will the product be?
