Let
$$\lim_{x \to a} f(x) = L$$
$$\lim_{x \to a} g(x) = M$$
Where $L$ and $M$ are finite reals.
Then I want to prove that
$$\lim_{x \to a} f(x) g(x) = LM$$
Let $\epsilon > 0$. We need a $\delta > 0$ such that for all $x$ we have $0 < |x-a| < \delta$ implying $|f(x)g(x) – LM| < \epsilon$.
Rearrange:
$$\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\
&=|g(x)(f(x)-L)+L(g(x)-M)|\\
&\le|g(x)||f(x)-L|+|L||g(x)-M| \\
&\lt|g(x)||f(x)-L|+(|L| + 1)|g(x)-M| < \epsilon\end{align}$$
Since the limits for $g(x)$ and $M$ approach the same value $M$, there exists a $\delta_1 > 0$ such that for all $x$, $0 < |x-a| < \delta_1$ implies $|g(x) – M| < \frac{\epsilon}{2(|L|+1)}$. Then:
$$\begin{align}|f(x)g(x)-LM| &\lt |g(x)||f(x)-L|+(|L|+1)|g(x)-M|\\
&<|g(x)||f(x)-L|+(|L|+1)\frac{\epsilon}{2(|L|+1)} \\
&=|g(x)||f(x)-L|+\frac{\epsilon}{2} = \epsilon \\
\end{align}$$
Since the limits for $f(x)$ and $L$ approach the same value $L$, there exists a $\delta_2 > 0$ such that for all $x$, $0 < |x-a| < \delta_2$ implies $|f(x) – L| < \frac{\epsilon}{2(|M|+1)}$. Then:
$$\begin{align}|f(x)g(x)-LM| &\lt |g(x)||f(x)-L|+\frac{\epsilon}{2} \\
&<|g(x)|\frac{\epsilon}{2(|M|+1)}+\frac{\epsilon}{2} = \epsilon \\
\end{align}$$
Now we prove $|g(x)| \leq |M|+1$:
$$|g(x)| = |g(x) – M + M| \leq |g(x) – M| + |M| \leq |M|+1$$
Subtracting $|M|$ from both sides, we see that:
$$|g(x) – M| \leq 1$$
Since the limits for $g(x)$ and $M$ approach the same value $M$, there exists a $\delta_3$ such that for all $x$, $0 < |x-a| < \delta_3$ implies $|g(x) – M| < 1$.
$$\begin{align}|f(x)g(x)-LM| &\lt |g(x)|\frac{\epsilon}{2(|M|+1)}+\frac{\epsilon}{2} \\
&< (|M|+1)\frac{\epsilon}{2(|M|+1)}+\frac{\epsilon}{2} \\
&= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon \\
\end{align}$$
This is true granted that we set $\delta = \min(\delta_1, \delta_2, \delta_3)$.
Is my proof correct and accurate? Have I actually proved the rule?
Best Answer
Rewriting your proof:
Let $|f(x) - L| < \epsilon_1$ for $|x-a| < \delta_1$
Let $|g(x) - M| < \epsilon_2$ for $|x-a| < \delta_2$
\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &=|g(x)(f(x)-L)+L(g(x)-M)|\\ &\le|g(x)||f(x)-L|+|L||g(x)-M| \\ &\lt|M|\epsilon_1+|L|\epsilon_2 \end{align}
Since $\epsilon_1$ and $\epsilon_2$ are arbitrarily small, the proof can end here I suppose.