# Thread: Continuous Problem

1. ## Continuous Problem

Suppose that f and g are defined and finite valued on an open interval I which contains a, that f is cont. at a, and that f(a) =/= 0. Then g is cont. at a if and only if fg is cont. at a.

2. Originally Posted by Math2010
Suppose that f and g are defined and finite valued on an open interval I which contains a, that f is cont. at a, and that f(a) =/= 0. Then g is cont. at a if and only if fg is cont. at a.
I mean, what have you done? Is $fg$ multiplication or composition?

3. fg is multiplication.

I know we are given $\forall \epsilon > 0, \exists \delta_f > 0, \delta_g > 0$ s.t $x,a \in A$

$|x - a| < \delta_f, |x - a| < \delta_g \forall x,a \in A \Rightarrow |f(x)-f(a)|<\epsilon, |g(x)-g(a)|<\epsilon$

I think I can simplify $|fg(x) - fg(a)|$ pretty easily to get it to equal $f(x)|g(x) - g(a)| + g(a)|f(x)-f(a)|$

I understand that $|g(x) - g(a)|$ and $|f(x)-f(a)|$ represent epsilon so we have $|fg(x) - fg(a)|\le f(x)\epsilon + g(a)\epsilon$ but now is where I really do not know what I need to do for f(x) and g(a). Thank you for the help.