Recent content by buZZ

Okay, abandon my initial approach. I think it'd be ten times easier if I show that 1/g(x) is continuous at a. Since g(a) is not zero, then g(x) has to bounded away from zero. Which means that 1/g(x) will not get infinitely large, but rather, will be bounded by something else. Now if only I...

that only seems useful if I'm trying to prove that f*g is continuous at a, but I'm lost as to see why that's useful for f/g

It doesn't seem that easy. Maybe I should back up. The fact that f and g are continuous at a mean that for |x-a|<delta, |f(x)-f(a)|<Something involving epsilon(1) |g(x)-g(a)|<something involving ep(2) By triangle inequality, |f(x)|< something involving E(1) + |f(a)| |g(x)|< something...

sorry about that, I didn't realize you could do that...I'm new here, And I need to use the rigorous definitions of continuity, so deriving it here from what I know about the bunches of limit theorems I know isn't sufficient.

you know...I could be going about it completely the wrong way too..

Homework Statement Okay, so if f and g are continuous functions at a, then prove that f/g is continuous at a if and only if g(a) # 0 Homework Equations Assuming to start off the g(a)#0, by the delta-epsilon definition of continuity, basically, We know that |f(x)| and |g(x)| are bounded...