# Explanation of the product rule for limits.

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 𝑓 and 𝑔 be real functions with $$\lim_{x\to 0}$$ 𝑓(𝑥) = 𝐹 and $$\lim_{x\to0}$$ 𝑔(𝑥) = 𝐺. Then the $$\lim_{x\to 0}$$𝑓(𝑥)𝑔(𝑥) exists and equals 𝐹𝐺.

Let 𝜀 > 0.

Then, there exists $$𝛿_1$$, $$𝛿_2$$, $$𝛿_3$$ such that

|𝑓(𝑥)−𝐹| < $$𝜀_2$$(1+|𝐺|) when 0 < |𝑥−$$𝑥_0$$| < $$𝛿_1$$ (1)

|𝑔(𝑥)−𝐺| < $$𝜀_2$$(1+|𝐹|) when 0 < |𝑥−$$𝑥_0$$| < $$𝛿_2$$ (2)

|𝑔(𝑥)−𝐺| < 1 when 0 < |𝑥−$$𝑥_0$$| < $$𝛿_3$$ (3)

According to the condition (3) we see that

|𝑔(𝑥)| = |𝑔(𝑥)−𝐺+𝐺| ≦ |𝑔(𝑥)−𝐺|+|𝐺| < 1+|𝐺| when 0 < |𝑥−$$𝑥_0$$| < $$𝛿_3$$.

Supposing then that, 0 < |𝑥 − $$𝑥_0$$| < min{$$𝛿_1$$,$$𝛿_2$$,$$𝛿_3$$}, and using (1) and (2)

Then

|𝑓(𝑥)𝑔(𝑥)−𝐹𝐺| = |𝑓(𝑥)𝑔(𝑥)−𝐹𝑔(𝑥)+𝐹𝑔(𝑥)−𝐹𝐺|

≦ |𝑓(𝑥)𝑔(𝑥)−𝐹𝑔(𝑥)|+|𝐹𝑔(𝑥)−𝐹𝐺|

= |𝑔(𝑥)|⋅|𝑓(𝑥)−𝐹|+|𝐹|⋅|𝑔(𝑥)−𝐺|

< (1+|𝐺|) $$𝜀_2$$ (1+|𝐺|)+(1+|𝐹|) $$𝜀_2$$ (1+|𝐹|) = 𝜀

• Your question is not clearly visible on Firefox on Android. Please use mathjax properly. – Paramanand Singh Nov 22 '19 at 3:35

## 1 Answer

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 $$|g(x)|<1+|G|$$.

• Ahhh that makes sense, thanks for the help. I am just getting used to using mathjax, feel free to make any edits to my post as you see fit. – Milton P. Nov 22 '19 at 3:42