I've been reading the product rule for limits and wondering if this proof is enough and if not, why.
Claim:
$f$ and $g$ are functions defined on an open interval $I$ with $c∈I$, with $x∈I$
$lim_{x\to c}f(x) = F, lim_{x\to c}g(x) = G,$
then
$(lim_{x\to c}f(x)lim_{x\to c}G(x))=lim_{x\to c}(f(x)g(x))$
My attempt at a proof:
$\forall \sqrt\epsilon>0\exists \delta_1>0$ such that if $|x-c|<\delta_1 \implies |f(x)-F| < \sqrt \epsilon$
and
$\forall \sqrt\epsilon>0\exists \delta_2>0$ such that if $|x-c|<\delta_2 \implies |g(x)-G| < \sqrt \epsilon$
let $\delta = min${$\delta_1,\delta_2$}
then $\forall |x-c| < \delta \implies=|(f(x)-F)(g(x)-G)|=|f(x)-F||g(x)-G| < \epsilon$
$\blacksquare$
Does this proof prove the claim? Many thanks.
Edit: misworded my question apologies!
Best Answer
I'll get you started with a technique that will help you select the $\delta$ you need.
$$|f(x)g(x)-FG| = |f(x)g(x)-f(x)G+f(x)G-FG|=|f(x)(g(x)-G)+G(f(x)-F)| \leq |f(x)||g(x)-G|+|G||f(x)-F|.$$
I'll leave it to you to choose $\delta$ that makes each component of the sum small enough.