I want to check my proof of the following statement:
if $\lim_{x \rightarrow a}f(x) = L_1$ and $\lim_{x \rightarrow a}g(x) = L_2$, then $\lim_{x \rightarrow a}(f \cdot g)(x) = L_1L_2$
I can't find any proof that is similar to mine, so I am curious whether my proof is valid.
Proof. Fix $\epsilon > 0 $. Let $\delta_1, \delta_2 > 0$ be such that $$0 < |x-a| < \delta_1 \implies |f(x) - L_1| < \sqrt{\epsilon}$$ and $$0 < |x-a| < \delta_2 \implies |g(x) - L_2| < \sqrt{\epsilon}$$ Let $\delta = \min\{\delta_1, \delta_2\}$. Then for any $x$ such that $0 < |x-a| < \delta$, we have \begin{align} |f(x)g(x) - L_1L_2| &\leq |f(x)g(x) - L_1L_2| + |- f(x)L_2 - L_1g(x) + 2L_1L_2| \\ &\leq |f(x) - L_1||g(x) - L_2| \qquad \text{(by triangle inequality)} \\ &< \epsilon \qquad \square\end{align}
I would appreciate any comment on this proof, thanks in advance.
Edit: For the second $\leq$ I used the following reasoning: \begin{align} |f(x)g(x) - L_1L_2| + |- f(x)L_2 - L_1g(x) + 2L_1L_2| &\leq |(f(x)g(x) - L_1L_2) + (- f(x)L_2 - L_1g(x) + 2L_1L_2)| \\ &= |f(x)g(x) - f(x)L_2 - L_1g(x) + L_1L_2| \\ &= |(f(x) - L_1)(g(x) - L_2)| \\ &= |f(x) - L_1||g(x) - L_2| \\ &< \sqrt{\epsilon}\sqrt{\epsilon} \\ &= \epsilon \end{align}
Unfortunately, your proof is invalid.
The key point is that you use what you call "triangle inequality" the wrong way around. For numbers $a, b \in \mathbb{R}$, the triangle inequality states that $|a+b| \leq |a| + |b|$. This is true. However, you use the converse statement, namely that $|a| + |b| \leq |a+b|$. This is not always true. For example, take $a=1$, $b=-1$. The left-hand side is $|1|+|-1| = 1+1 = 2$, while the right-hand side is $|1+(-1)|=|0|=0$.