From this 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.
I thought we are supposed to find $\delta > 0$ such that $|f(x)g(x)-LM| < \varepsilon$ holds for all $\varepsilon > 0$. It may be that $\delta = \min\{\delta_1, \delta_2\}$. What I mean is that there should be something like a proof of the desired limit, where it starts from $\delta$ which inevitably ends with $\varepsilon$.
For example, if $\lim\limits_{x \to a}f(x) = L$ and $\lim\limits_{x \to a}g(x) = M$, then there exists $\delta_1$ and $\delta_2$ such that $|f(x) – L| < \frac{\varepsilon}{2}$ holds whenever $0 < |x – a| < \delta_1$ and $|g(x) – M| < \frac{\varepsilon}{2}$ holds whenever $0 < |x – a| < \delta_2$ holds, for $\varepsilon > 0$. Now, we can let $\delta = \min\{\delta_1, \delta_2\}$ such that $\left|\left(f(x) + g(x)\right) – (L + M)\right| < \epsilon$ whenever $0 < |x – a| < \delta$. The remaining part is the proof of the desired limit, which is
\begin{align*}
|(f(x) + g(x)) – (L + M)| &= |(f(x) – L) + (g(x) – M)| \\
|(f(x) + g(x)) – (L + M)| &\leq |f(x) – L| + |g(x) – M| \\
|(f(x) + g(x)) – (L + M)| &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\
|(f(x) + g(x)) – (L + M)| &< \varepsilon
\end{align*}
Going back to the answer, I think $|g(x)|$ became $M$ as $g(x)$ approaches $M$. The same thing applies to $f(x)$ and $L$. However, I don't understand how $|M|\varepsilon_1 + |L|\varepsilon_2$ can be $\varepsilon$. Can someone explain this part, and why the proof can end on the inequality $|f(x)g(x) – LM| < |M|\varepsilon_1 + |L|\varepsilon_2$?
Best Answer
Since $L$ and $M$ are fixed, and $\epsilon_1$ and $\epsilon_2$ can be chosen arbitrarily at the beginning of the proof, they can be chosen so that $|M| \epsilon_1 + |L| \epsilon_2 < \epsilon$ (doesn't need to be equal to $\epsilon$) in order to conclude that $|f(x)g(x) - LM| < \epsilon$.
If you really want it to be equal to $\epsilon$, you could choose $\epsilon_1 = \frac{\epsilon}{2 |M|}$ and $\epsilon_2 = \frac{\epsilon}{2 |L|}$, for example.