# Proving the product rule for limits

Proof: $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$

Let $L_1 = \lim_{x \to a} f(x)$ and $L_2 = \lim_{x \to a} g(x)$. We assume $L_1, L_2 \neq 0$ (for scenarios where $L_1$ or $L_2$ are $0$, the result is trivially true).

Let $\epsilon > 0$. We need a $\delta$ such that $|f(x)g(x) - L_1L_2| < \epsilon$ whenever $0 < |x-a| < \delta$.

We can rewrite

\begin{align}|f(x)g(x) - L_1L_2| &= |(f(x)-L_1)(g(x)-L_2) + L_1(g(x)-L_2) + L_2(f(x)-L_1)| \\ &\leq |f(x)-L_1||g(x)-L_2| + |L_1||g(x)-L_2| + |L_2||f(x)-L_1| \\ &< \epsilon\end{align}

by triangle inequality.

Let $\delta_1 > 0$ such that $|f(x)-L_1| < \sqrt{\frac{\epsilon}{3}}$ whenever $0 < |x-a| < \delta_1$.

Let $\delta_2 > 0$ such that $|g(x)-L_2| < \sqrt{\frac{\epsilon}{3}}$ whenever $0 < |x-a| < \delta_2$.

Let $\delta_3 > 0$ such that $|g(x)-L_2| < \frac{\epsilon}{3|L_1|}$ whenever $0 < |x-a| < \delta_3$.

Let $\delta_4 > 0$ such that $|f(x)-L_1| < \frac{\epsilon}{3|L_2|}$ whenever $0 < |x-a| < \delta_4$.

Suppose that $0 < |x-a| < \delta$ where $\delta = \min(\delta_1, \delta_2, \delta_3, \delta_4)$. Then we have:

\begin{align}|f(x)-L_1||g(x)-L_2| &+ |L_1||g(x)-L_2| + |L_2||f(x)-L_1| \\&< \sqrt{\frac{\epsilon}{3}} \cdot \sqrt{\frac{\epsilon}{3}} + |L_1|\frac{\epsilon}{3|L_1|} + |L_2|\frac{\epsilon}{3|L_2|} \\&= \epsilon \end{align}

Is this a correct proof?

your proof of the essential point seems flawless. one might only raise one small quibble: the statement of what you are proving is a little imprecise. suppose $f(x)=x$ and $g(x)=\frac1x$. then $\forall x.f(x)g(x)=1$ so
$$\lim_{x \to 0} f(x)g(x) = 1$$ but what is $\lim_{x\to 0}f(x) \lim_{x \to 0}g(x)$?