# 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 π and π be real functions with $$\lim_{x\to 0}$$ π(π₯) = πΉ and $$\lim_{x\to0}$$ π(π₯) = πΊ. Then the $$\lim_{x\to 0}$$π(π₯)π(π₯) exists and equals πΉπΊ.

Let π > 0.

Then, there exists $$πΏ_1$$, $$πΏ_2$$, $$πΏ_3$$ such that

|π(π₯)βπΉ| < $$π_2$$(1+|πΊ|) when 0 < |π₯β$$π₯_0$$| < $$πΏ_1$$ (1)

|π(π₯)βπΊ| < $$π_2$$(1+|πΉ|) when 0 < |π₯β$$π₯_0$$| < $$πΏ_2$$ (2)

|π(π₯)βπΊ| < 1 when 0 < |π₯β$$π₯_0$$| < $$πΏ_3$$ (3)

According to the condition (3) we see that

|π(π₯)| = |π(π₯)βπΊ+πΊ| β¦ |π(π₯)βπΊ|+|πΊ| < 1+|πΊ| when 0 < |π₯β$$π₯_0$$| < $$πΏ_3$$.

Supposing then that, 0 < |π₯ β $$π₯_0$$| < min{$$πΏ_1$$,$$πΏ_2$$,$$πΏ_3$$}, and using (1) and (2)

Then

|π(π₯)π(π₯)βπΉπΊ| = |π(π₯)π(π₯)βπΉπ(π₯)+πΉπ(π₯)βπΉπΊ|

β¦ |π(π₯)π(π₯)βπΉπ(π₯)|+|πΉπ(π₯)βπΉπΊ|

= |π(π₯)|β|π(π₯)βπΉ|+|πΉ|β|π(π₯)βπΊ|

< (1+|πΊ|) $$π_2$$ (1+|πΊ|)+(1+|πΉ|) $$π_2$$ (1+|πΉ|) = π

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

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