The following proof of the product rule for limits was given in my analysis class, and seems to be the standard proof for this property. However, could someone explain why (3) is necessary? I am having a hard time understanding why that is needed, why aren't (1) and (2) sufficient without (3)? Thanks in advance!
Let and be real functions with $\lim_{x\to 0}$ () = and $\lim_{x\to0}$ () = . Then the $\lim_{x\to 0}$()() exists and equals .
Let > 0.
Then, there exists $_1$, $_2$, $_3$ such that
|()−| < $_2$(1+||) when 0 < |−$_0$| < $_1$ (1)
|()−| < $_2$(1+||) when 0 < |−$_0$| < $_2$ (2)
|()−| < 1 when 0 < |−$_0$| < $_3$ (3)
According to the condition (3) we see that
|()| = |()−+| ≦ |()−|+|| < 1+|| when 0 < |−$_0$| < $_3$.
Supposing then that, 0 < | − $_0$| < min{$_1$,$_2$,$_3$}, and using (1) and (2)
Then
|()()−| = |()()−()+()−|
≦ |()()−()|+|()−|
= |()|⋅|()−|+||⋅|()−|
< (1+||) $_2$ (1+||)+(1+||) $_2$ (1+||) =
I think there are some typos in your proof. You have $|g(x)|<1+|G|$, and you want to have the bounds $|f(x)-F|<\epsilon /2(1+|G|)$ and $|g(x)-G|<\epsilon /2(1+|F|)$, then $$ \begin{align*} |f(x)g(x)-FG|&=|f(x)g(x)-Fg(x)+Fg(x)-FG|\\ &\leqslant |f(x)g(x)-Fg(x)|+|Fg(x)-FG|\\ &=|g(x)||f(x)-F|+|F||g(x)-G|\\ &\leqslant (1+|G|)\frac{\epsilon }{2(1+|G|)}+|F|\frac{\epsilon }{2(1+|F|)}\\ &\leqslant\epsilon \end{align*} $$ when $0<|x|<\delta $ for suitable $\delta >0$ (note that $x_0=0$ in your example). In your proof $(3)$ is needed to set the bound $|g(x)|<1+|G|$.