# Proof check: limit of product is equal to product of limits

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}

• In the final calculation, the second $\le$ step seems unjustified to me. Commented Apr 25, 2023 at 6:25
• @GregMartin Hi Greg, thanks for your feedback. I edited my question to include my reasoning for the second $\leq$. Does it make sense? Commented Apr 25, 2023 at 6:36
• I've never seen an epsilon-delta proof using a $\sqrt{\epsilon}$ bound on errors, for this or anything else. The usual approach uses $fg-L_1L_2=f(g-L_2)+(f-L_1)L_2$ to prove $\epsilon_i<\min\left\{1,\,\frac{\epsilon}{2(1+|L_{3-i}|)}\right\}$ for $i\in\{1,\,2\}$ suffices.
– J.G.
Commented Apr 25, 2023 at 7:57

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 $$|1+(-1)|=|0|=0$$.