I'm looking for some help with proving that if f and g are continuous then so is f/g. I need to use the ε and Δ definition of continuity
( │ x - a│< Δ implies │f(x) – f(a)│< ε) to prove this.
Of course, we must assume $\displaystyle g(a)\neq 0$, and then :
$\displaystyle \forall,\epsilon>0\,\,\exists\,\delta_1>0\,\,s.t.\ ,\,|x-a|<\delta_1\,\Longrightarrow\,|f(x)-f(a)|<\frac{e}{2g(a)}\,\,\,and\,$ $\displaystyle \exists\,\,\delta_2>0\,\,\,s.t.\,\,|x-a|<\delta_2\,\Longrightarrow\,|g(x)-g(a)|<\frac{2}{2f(a)}$
We also have that $\displaystyle |g(a)-g(x)|<\frac{\epsilon}{2f(a)}\Longrightarrow\,|g(x) |>|g(a)|-\frac{\epsilon}{2f(a)}$ , thus for $\displaystyle |x-a|<min(\delta_1,\delta_2)$:
$\displaystyle \left|\frac{f(x)}{g(x)}-\frac{f(a)}{g(a)}\right|=\left|\frac{f(x)g(a)-f(a)g(a)+f(a)g(a)-f(a)g(x)}{g(x)g(a)}\right|$ $\displaystyle \leq\frac{\left|f(x)-f(a)\right||g(a)|+\left|g(a)-g(x)\right||f(a)|}{|g(x)g(a)|}$ $\displaystyle <\frac{\frac{\epsilon}{2g(a)}|g(a)|+\frac{\epsilon }{2f(a)}|f(a)|}{|g(x)g(a)|}=\frac{\epsilon}{K}$ , where K is a
constant (as g(x) is as close to g(a) as we want if x close enough to a).
Tonio