Algebraic closure of limit existence? Problem: If $\lim_{x\to a}[f(x)+g(x)]=2$ and $\lim_{x\to a}[f(x)-g(x)]=1$, then find $\lim_{x\to a}[f(x)g(x)]$. 
I came up with a way that I know works, but someone else showed me a more interesting way of going about it...
We know
$$
\lim_{x\to a}[f(x)+g(x)]^2=(\lim_{x\to a}[f(x)+g(x)])^2=4=\lim_{x\to a}(f(x)^2+2f(x)g(x)+g(x)^2)\tag{1}
$$
and
$$
\lim_{x\to a}[f(x)-g(x)]^2=(\lim_{x\to a}[f(x)-g(x)])^2=1=\lim_{x\to a}(f(x)^2-2f(x)g(x)+g(x)^2).\tag{2}
$$
By subtracting $(2)$ from $(1)$, we get
$$
\lim_{x\to a}4[f(x)g(x)]=3
$$
or 
$$
\lim_{x\to a}[f(x)g(x)]=\frac{3}{4}.
$$
Even though the result of $3/4$ is correct, is this not a logical violation? Were the limit laws used properly? It doesn't seem as though we should be able to subtract $(2)$ from $(1)$ in the way it was. Even if this were valid (is it?), then would it be correct to say $\lim_{x\to a}4[f(x)g(x)]=3$ is the same as $4\lim_{x\to a}[f(x)g(x)]$? Wouldn't this be using another limit law that assumed the existence of $\lim_{x\to a}[f(x)g(x)]$?  
 A: It's very good that you should be dubious and careful!
But we have a theorem that if
$\lim f (x)= k $ and $\lim g (x)=j $ then $\lim cf (x)=ck $ and $\lim[ f (x)+g (x)]=k+j $ and $\lim f (x)g (x)=kj $ and if $k\ne 0$ then $\lim \frac 1 {f (x)}=\frac 1k $.
So everything there is perfectly valid!
....
Which isn't to say that in $\lim f (x)g (x)= k $ that $\lim f (x) $ or $\lim g (x)$ actually exist.  
... oh, I see your concern!
We know $\lim h(x)=k\implies \lim ch (x)=ck $ but that wouldn't mean $\lim ch (x)=k\implies \lim h (x)=\frac kc $ if we don't know the limit exists in the first place.
But it does!
Let $j (x)=ch (x) $ then we have $\lim j (x) =\lim ch (x)=k $.  So $\lim \frac 1c j (x) =\frac 1c k $.
===
Redo:
Everything is valid.
Let $j (x)=f (x)+g (x) $ let $k (x)=f (x)-g (x) $.  Let $h (x)=4f (x)g (x)=j^2 (x)-k^2 (x) $.
$\lim j (x)=2$; $\lim k (x)=1$ so $\lim h (x)=\lim( j^2 (x)-k^2 (x))=2^2-1^2=3$
So $\lim \frac 14 h (x)=\lim f (x)g (x)=\frac 14 *3=\frac 34$.
This is perfectly legitimate even if $\lim f (x) $ or $\lim g (x)$ don't exist.
A: hint
use
$$ab=\frac {(a+b)^2-(a-b)^2}{4} $$
A: There is a slight generalization of the algebra of limits which is used implicitly almost everywhere, but not mentioned explicitly in most textbooks:
We have $$\lim_{x\to a} f(x) \pm g(x) = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x) $$ provided at least one of the limits on right exists and $$\lim_{x\to a} f(x) g(x) =\lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x) $$ provided at least one of the limits on right exists and is nonzero.
The above rules allow us to apply algebra of limits without worrying too much about existence of limits of sub-expressions while evaluating limit of a complicated expression. The questions of existence of limits get settled during the process of evaluation itself.
Your example is simpler and it does not really require the above generalization, the usual rules of limits mentioned in textbooks suffice. Each step in your calculation is validated by a specific rule and your concern about the validity of such steps is more than what is usually necessary. Thus we use the product rule when we square the limits, then the rule for subtraction of limits is used and finally the quotient rule is used (although here the quotient is by a constant function $4$).
