Prove limit of product of functions is product of limit of functions I am given that limits of $f(x)$ and $g(x)$ exist. Lets suppose they are $L$ and $M$ respectively. So, I am given that
$$\lim_{x \to a} f(x) = L \;\; \lim_{x \to a} g(x) = M$$
I have to prove that
$$\lim_{x \to a} \; \bigl[ f(x) g(x) \bigr] = \lim_{x \to a} f(x) \bullet\lim_{x \to a} g(x) $$
I have to do a $\varepsilon-\delta$ proof here. So, I write what I have been given using qualifiers as follows.
$$\forall \varepsilon >0 \; \exists \delta > 0\; \forall x \in D_1 \; \bigl[ 0 < |x-a| < \delta \Rightarrow |f(x) - L| < \varepsilon \bigr ] \cdots\cdots (1)$$
$$\forall \varepsilon >0 \; \exists \delta > 0\; \forall x \in D_2 \; \bigl[ 0 < |x-a| < \delta \Rightarrow |g(x) - M| < \varepsilon \bigr ] \cdots\cdots (2)$$
And I have to prove the following
$$\forall \varepsilon >0 \; \exists \delta > 0\; \forall x \in D \; \bigl[ 0 < |x-a| < \delta \Rightarrow |f(x)g(x) - LM| < \varepsilon \bigr ] \cdots\cdots (3)$$
Here $D = D_1 \cap D_2$. So, I let $\varepsilon >0$ be some arbitrary number. Now, I have $|L| \geqslant 0$ and $|M| \geqslant 0$. So, following inequalities hold
$$ \frac{\varepsilon}{3(|M| + 1)} > 0$$
$$ \frac{\varepsilon}{3(|L| + 1)} > 0$$
So, using these in place of $\varepsilon$ in equations $(1)$ and $(2)$, I get that, there exist $\delta_1 > 0$ and $\delta_2 > 0$ such that
$$ \forall x \in D_1 \; \biggl[ 0 < |x-a| < \delta_1 \Rightarrow |f(x) - L| < \frac{\varepsilon}{3(|M| + 1)} \biggr ]  \cdots\cdots (4)$$
$$ \forall x \in D_2 \; \biggl[ 0 < |x-a| < \delta_2 \Rightarrow |g(x) - M| < \frac{\varepsilon}{3(|L| + 1)} \biggr ] \cdots\cdots (5) $$
Since $\varepsilon > 0$, I also have $\varepsilon^{1/2} > 0$ and $\frac{\varepsilon^{1/2}}{3} > 0$. So, there exist $\delta_3 > 0$ and $\delta_4 > 0$ such that
$$ \forall x \in D_1 \; \biggl[ 0 < |x-a| < \delta_3 \Rightarrow |f(x) - L| <  \frac{\varepsilon^{1/2}}{3}\biggr ]  \cdots\cdots (6)$$
$$ \forall x \in D_2 \; \biggl[ 0 < |x-a| < \delta_4 \Rightarrow |g(x) - M| < \varepsilon^{1/2} \biggr ] \cdots\cdots (7) $$
Now, let $\delta = \text{min}(\delta_1, \delta_2, \delta_3, \delta_4)$. Its obvious that $\delta >0$. Since, I have to prove $(3)$, I let $x \in D$ some arbitrary element. So, I have that $x \in D_1$ and $x \in D_2$. Also, suppose $ 0 < |x-a| < \delta$. It follows that
$$ 0 < |x-a| < \delta_1  $$
$$ 0 < |x-a| < \delta_2  $$
$$ 0 < |x-a| < \delta_3  $$
$$ 0 < |x-a| < \delta_4  $$
Now using equations $(4),(5),(6),(7)$ with Modus ponens, we can deduce the following
$$ |f(x) - L| < \frac{\varepsilon}{3(|M| + 1)} \\ 
 |g(x) - M| < \frac{\varepsilon}{3(|L| + 1)} \\
 |f(x) - L| <  \frac{\varepsilon^{1/2}}{3} \\
 |g(x) - M| < \varepsilon^{1/2} 
 $$
Now, using triangle inequality, I can write
$$ |f(x)g(x) - LM| \leqslant |f(x) - L| \;|g(x) - M|  + |M| \;|f(x) - L| + |L| \;|g(x) - M|  $$
Using the inequalities already deduced, I get the following
$$ |f(x)g(x) - LM| < \frac{\varepsilon^{1/2}}{3}\cdot \varepsilon^{1/2}  + |M| \left[\frac{\varepsilon}{3(|M| + 1)}\right] + |L| \left[\frac{\varepsilon}{3(|L| + 1)} \right]\cdots\cdots(8)$$
Now, I also have following inequalities,
$$ 0 \leqslant \frac{|M|}{|M| + 1} < 1 $$
$$ 0 \leqslant \frac{|L|}{|L| + 1} < 1 $$
Since $\varepsilon / 3 > 0$, it follows that
$$ |M| \left[\frac{\varepsilon}{3(|M| + 1)}\right] < \frac{\varepsilon}{3} $$
$$ |L| \left[\frac{\varepsilon}{3(|L| + 1)}\right] < \frac{\varepsilon}{3} $$
So, inequality $(8)$ now becomes
$$ |f(x)g(x) - LM| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} $$
$$ |f(x)g(x) - LM| < \varepsilon $$
Since $x \in D$ and $\varepsilon > 0 $ were arbitrary to begin with, equation $(3)$ follows. So, it follows that
$$\lim_{x \to a} \; \bigl[ f(x) g(x) \bigr] = \lim_{x \to a} f(x) \bullet\lim_{x \to a} g(x) $$
Is the proof valid ?
Thanks
 A: Not an answer but a general note:
I don't have the energy/time to read through your whole proof, but it might be worth defining
$$
F(x) = f(x)- L \\
G(x) = g(x) - M.
$$
Then you have $\lim_{x \to a} F(x) = 0$, and similarly for $G$. If you can prove something about the limit of $FG$, then you'll pretty quickly be able to show what you need about $fg$ as well.
This trick --- making as many things zero as possible --- is often useful. In fact, you can go a step further and let
$$
F(x) = f(x+a) - L\\
G(x) = g(x+a) = M
$$
and then prove something about the limit of $FG$ at $x = 0$, which makes various things even simpler.
A: I don't actually know what is the meaning of $D_1$. Besides the def of $D_i$, the proof should be right but too abundant, I can see that you put too much emphasis on proof of $\epsilon^2$. It's better to broaden the judge of primary $\epsilon$, as $$\forall\epsilon >0, \exists \delta_1>0\quad s.t.|x-a|<\delta_1\rightarrow |f(x)-L|<\frac {\epsilon}{100(M+L)}$$the same as $\delta_2$,then you set $\delta$ as $min${$\delta_1,\delta_2$}, the proof maybe simplier and clearer
