Proof of Product Rule of Limits How can I prove the product rule of limits? I want to prove this but I am stuck while proving it. Please help me. 
I mean:
$$\lim\limits_{x\to c}\  (f(x)\cdot g(x)) = \left [\lim_{x\to c}\ f(x) \right] \cdot \left [\lim_{x\to c} \ g(x) \right] $$
 A: HINT $\: $ It's simple if $\rm\:f\to 0\:$ or $\rm\:g\to 0\:.\:$ By linearity we can reduce the general case to a sum of such simple cases as follows. Suppose $\rm\:f\to a,\ g\to b\:.\:$ Then $\rm\ f_0 := f-a \to 0,\ \ g_0 := g-b\to 0\:,\ $ hence
$$\rm f\:g - ab\ = \ (f_0+a)\:(g_0+b) - a\:b\ =\ f_0\ g + a\ g_0\ \to\ 0 $$
since the final term above is a sum of said cases that are simply proved to $\to 0\:.\ $  An analogous proof works also for many other product rules, e.g. for congruences in  number theory we have 
Congruence Product Rule $\rm\ \ f\equiv a,\ g\equiv b\ \Rightarrow\ f\:g\equiv ab\ \ (mod\ m)$
Proof $\rm\:\ \ m\ |\ f-a,\ g-b\:\ \Rightarrow\:\ m\ |\ (f-a)\ g + a\ (g-b)\ =\ f\:g - ab\quad $  QED
The "simple case" used above is $\rm\ h\equiv 0\ \Rightarrow\ c\:h\equiv 0\ $ i.e. $\rm\ m\ |\ h\ \Rightarrow\ m\ |\ c\:h$
When you study ring theory you will learn that the fundamental algebraic structure at the heart of the above proofs is the notion of an ideal in a ring, or a module over a ring. Namely, above, the elements that $\to 0\:$ or are $\equiv 0$ resp., form a subgroup (i.e. are closed under subtraction) and are further closed under multiplication by an arbitrary ring element ("scaling"). This fundamental linear structure  is a ring-theoretic generalization of the structure of a vector space over a field. However, since the scalars need only be a ring (vs. a field), the theory is a bit more complicated, due in large part to the fact that one can no longer always cancel or invert nonzero scalars.
