If $f$ is continuous at $0$, and $g$ is not continuous at $0$, then $fg$ will not be continuous at $0$? I know this seems very basic and obvious but I am just now learning about continuity and limits at school. So in one of my classes, under the theorem category it was put that if $f$ is continuous at point $a$ and $g$ is continuous at point $a$ then $fg$ will be continuous at point $a$.
So then that means that if $f$ is continuous at $0$ but $g$ is not continuous at $0$, then $fg$ will not be continuous at $0$.
Then I found a function $f$ where it is continuous at $0$, $f(x)$=x.
And then I found a function $g$ where it is not continuous at $0$, $g(x)=1/x$.
However if you take $fg$ you will have $x × 1/x$ which is $x/x=1$. But isn't $fg=1$ continuous at $0$ even when $g$ is not? Did I understand the theorem wrong?
 A: This is actually a very nice observation. Let me explain by way of the similar rule for addition.
Here is a true fact: Suppose $f$ is continuous. If $g$ is continuous then $f+g$ is continuous.
The equally true contrapositive is: Suppose $f$ is continuous. If $f+g$ is not continuous then $g$ is not continuous.
And, we can replace $f$ by $-f$ (which preserves continuity) and then $g$ by $f+g$ to reach another true statement: Suppose $f$ is continuous. If $g$ is not continuous then $f+g$ is not continuous.
This looks like the inverse of the original implication (which, as Lee Mosher commented, is not automatically equivalent to the original implication), but it turns out to be true because of some symmetries in addition/subtraction of functions.
Now let's try the same procedure with this true fact: Suppose $f$ is continuous. If $g$ is continuous then $fg$ is continuous.
Its equally true contrapositive is: Suppose $f$ is continuous. If $fg$ is not continuous then $g$ is not continuous.
It seems like we could do the same thing: replace $f$ by $1/f$ and then $g$ by $fg$, which would yield the statement in the original post—supposing $f$ is continuous, then if $fg$ is not continuous then $g$ is not continuous.
However, replacing $f$ by $1/f$ is not quite as straightforward as replacing $f$ by $-f$. Anytime $f$ is continuous, $-f$ is automatically continuous as well. However, when $f$ is continuous, $1/f$ is only continuous when $f$ is nonzero.
So when we replace $f$ by $1/f$, we have to add the assumption that $f$ is nonzero to make the argument work. The correct deduction is: Suppose $f$ is continuous and nonzero. If $g$ is not continuous then $fg$ is not continuous.
So this isn't a case of a logical fallacy (we're not asserting the inverse of an implication)—the new statement only looks like the inverse because of some symmetries of multiplication/division. But this explicit derivation does point out why we need the assumption that $f$ is nonzero, as the OP's example shows.
A: You can not give any statements about the continuity of $g(x)$ at $x=0$, since $g(x)$ isn't even defined at $x=0$.
That means you cant say that it is continuous or not continuous at 0, it is just not existing at 0.
But to your original question. It is true that fg is continuous if f and g are continuos. But only because one of them is not continuous, the product still can be.
For example $f(x) = x$ and $g(x)= \begin{cases} 0 && x=0  \\ \sin(1/x) && \text{else}
\end{cases}$
Then $fg = x \sin(1/x)$ and this function indeed is continuous
