Proving that the limit of a product is the product of the limits Here's the theorem I'm trying to prove:
Let $f$ and $g$ be functions and $x_0 \in \mathbb{R}$. If $\lim_{x \to x_0} f(x) = L$ and $\lim_{x \to x_0} g(x) = M$, then $\lim_{x \to x_0} f(x) \cdot g(x) = LM$. 

Proof Attempt:
By a previously proven result, we have:
$$\lim_{x \to x_0} f(x) = L \iff f(x) = L + \alpha(x) \land \lim_{x \to x_0} \alpha(x) = 0$$
$$\lim_{x \to x_0} g(x) = M \iff g(x) = M + \beta(x) \land \lim_{x \to x_0} \beta(x) = 0$$
So, consider the product $f(x)g(x)$. This gives us the following equality:
$$f(x)g(x) = [L + \alpha(x)][M + \beta(x)] = LM + M\alpha(x) + L\beta(x) + \alpha(x)\beta(x)$$
$$\implies |f(x)g(x)-LM| = |M\alpha(x) + L\beta(x) + \alpha(x)\beta(x)|$$
$$\implies |f(x)g(x)-LM| \leq |L||\beta(x)|+|M||\alpha(x)|+|\alpha(x)\beta(x)|$$
By a previously proven result, we know that $\lim_{x \to x_0} \alpha(x)\beta(x) = 0$. So, let $\epsilon>0$ and $\delta_1,\delta_2,\delta_3>0$ be numbers such that:
$$0 < |x-x_0| < \delta_1 \implies |\alpha(x)\beta(x)| < \frac{\epsilon}{|L|+|M|+1}$$
$$0 < |x-x_0| < \delta_2 \implies |\alpha(x)| < \frac{\epsilon}{|L|+|M|+1}$$
$$0 < |x-x_0| < \delta_3 \implies |\beta(x)| < \frac{\epsilon}{|L|+|M|+1}$$
Let $\delta = min\{\delta_1,\delta_2,\delta_3\}$. Then, we have:
$$|f(x)g(x)-LM| \leq |L||\beta(x)| + |M||\beta(x)| + |\alpha(x)\beta(x)| < \frac{|L|\epsilon+|M|\epsilon+\epsilon}{|L|+|M|+1} < \epsilon$$
That shows that there does exist a $\delta>0$ such that the above inequality holds for any $\epsilon>0$. This proves the desired assertion. 
Could anyone check if my argument above works or not? If it doesn't, how do I fix it?
 A: Your proof sounds good. I provide another way to approach it just for the sake of curiosity.
Lemma 1
Let us assume that $\lim f(x) = L$ and $\lim g(x) = M$. Then $\lim(f(x) + g(x)) = L + M$.
Proof
According to the definition of limit, for every $\varepsilon/2 > 0$, there are $\delta_{1} > 0$ and $\delta_{2} > 0$ such that
\begin{align*}
\begin{cases}
0 < |x - x_{0}| < \delta_{1}\\\\
0 < |x - x_{0}| < \delta_{2}
\end{cases} \Longrightarrow
\begin{cases}
|f(x) - L| < \varepsilon/2\\\\
|g(x) - M| < \varepsilon/2
\end{cases}
\end{align*}
Consequently, for every $\varepsilon > 0$ and $\delta = \min\{\delta_{1},\delta_{2}\}$, one has that
\begin{align*}
0 < |x - x_{0}| < \delta \Longrightarrow |f(x) + g(x) - L - M| \leq |f(x) - L| + |g(x) - M| < \varepsilon
\end{align*}
and we have prove that $\lim(f(x) + g(x)) = L + M$
Lemma 2
Let us assume that $\lim f(x) = L$ and $k\in\textbf{R}$. Then $\lim kf(x) = kL$.
Proof
The case when $k = 0$ is trivial. So we shall consider that $k\neq 0$.
According to the definition of limits, for every $\varepsilon/|k| > 0$, there is a $\delta > 0$ such that
\begin{align*}
0 < |x - x_{0}| < \delta \Longrightarrow |f(x) - L| < \varepsilon/|k| \Longrightarrow |kf(x) - kL| < \varepsilon
\end{align*}
and we have proven that $\lim kf(x) = kL$.
Lemma 3
If $\displaystyle\lim_{x\rightarrow x_{0}}f(x) = L$ and $\displaystyle\lim_{y\rightarrow L}g(y) = g(L)$, then $\displaystyle\lim_{x\rightarrow x_{0}}g(f(x)) = g(L)$
Proof
According to the definition of limit, for every $\varepsilon > 0$, there exists a $\delta_{1} > 0$ such that
\begin{align*}
0 < |y - L| < \delta_{1} \Longrightarrow |g(y) - g(L)| < \varepsilon
\end{align*}
Similarly, for every $\delta_{1} > 0$, there exists a $\delta > 0$ such that
\begin{align*}
0 < |x - x_{0}| < \delta \Longrightarrow |f(x) - L| < \delta_{1}
\end{align*}
If we make the substitution $y = f(x)$, we conclude that, for every $\varepsilon > 0$, there exists a $\delta > 0$ such that the following relation holds
\begin{align*}
0 < |x - x_{0}| < \delta \Longrightarrow |g(f(x)) - g(L)| < \varepsilon
\end{align*}
from whence we conclude that $\displaystyle\lim_{x\rightarrow x_{0}}g(f(x)) = g(L)$, just as desired.
Lemma 4
The function $f(x) = x^{2}$ satisfies
\begin{align*}
\lim_{x\rightarrow x_{0}}x^{2} = x^{2}_{0}
\end{align*}
Proof
Assuming that $|x - x_{0}| < \delta$, we have that
\begin{align*}
|x^{2} - x^{2}_{0}| = |x - x_{0}||x + x_{0}| < \delta|x + x_{0}| < \delta(|x| + |x_{0}|) \leq \delta(\delta + 2|x_{0}|)
\end{align*}
Hence, for every $\varepsilon > 0$, there exists a $\delta > 0$ such that
\begin{align*}
0 < |x - x_{0}| < \delta \Longrightarrow |x^{2} - x^{2}_{0}| < \varepsilon
\end{align*}
This is because, for every $\varepsilon > 0$, the quadratic equation
\begin{align*}
\delta^{2} + 2|x_{0}|\delta - \varepsilon = 0
\end{align*}
does always have a positive root.
Proposition
Given that the product $f(x)g(x)$ can be rewritten as
\begin{align*}
f(x)g(x) = \frac{[f(x) + g(x)]^{2} - [f(x)]^{2} - [g(x)]^{2}}{2}
\end{align*}
the previous results ensure that
\begin{align*}
\lim f(x)g(x) = LM
\end{align*}
Hopefully this helps.
