# Confusion on proof of limit laws epsilon delta

at https://www.youtube.com/watch?v=9tYUmwvLyIA from 35:33 to 39:33, Herb Gross says: \begin{align} f(x) &= L+ [f(x)-L] \\ g(x) &= M+ [g(x)-M] \end{align} Multiplying these 2, we get: \begin{align} f(x)g(x) &= LM + L[g(x) -M] + M[f(x) -L] +[f(x) -L][g(x)-M] \\ f(x)g(x) - LM &= L[g(x) -M] + M[f(x) -L] +[f(x) -L][g(x)-M] \end{align} From here we can say: \begin{align*} |f(x)g(x) - LM| &= |L[g(x) -M] + M[f(x) -L] +[f(x) -L][g(x)-M]| \\ &\leq |L[g(x) -M]| + |M[f(x) -L]| +|[f(x) -L][g(x)-M]| \end{align*} From here we can impose: \begin{align} |L[g(x) -M]| &< \frac{\epsilon}{3} \\ |M[f(x) -L]| &<\frac{\epsilon}{3} \\ |f(x) -L] &< \sqrt{\frac{\epsilon}{3}} \\ |g(x)-M| &< \sqrt{\frac{\epsilon}{3}} \end{align} My problem is: \begin{align} |L[g(x) -M]| &< \frac{\epsilon}{3} \implies|g(x) -M| &< \frac{\epsilon}{3|L|} \\ |M[f(x) -L]| &<\frac{\epsilon}{3} \implies |f(x) -L| <\frac{\epsilon}{3|M|}\\ |f(x) -L] &< \sqrt{\frac{\epsilon}{3}} \\ |g(x)-M| &< \sqrt{\frac{\epsilon}{3}} \end{align} So does that mean: $$|f(x) -L] < \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|} \}$$ $$|g(x) -M] < \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|} \}$$

Is this how it would look like in a proof: Given $$\epsilon > 0$$ Let $$\epsilon_1 =\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|} \}$$ then $$\exists \delta_1 >0$$ such that: $$0<|x-a|<\delta_1 \implies |f(x) - L| < \epsilon_1$$ Let $$\epsilon_2 =\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|} \}$$ then $$\exists \delta_1 >0$$ such that: $$0<|x-a|<\delta_2 \implies |g(x) - M| < \epsilon_2$$ then let $$\delta \leq \min\{\delta_1,\delta_2\}$$ \begin{align*} 0<|x-a|<\delta &\implies |L||g(x) -M| + |M||f(x) -L| +|f(x) -L||g(x)-M| \\ &< |L|\times\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|} \} + |M|\times\min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|} \} + \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|L|} \} \times \min \{ \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|M|} \}\\ &< \epsilon \end{align*} This last part is not mentioned in the video is it all correct?

That is needlessly complicated.

(1). Prove that if $$A$$ is a constant and $$\lim_{x\to y}u(x)=V$$ then $$\lim_{x\to y}A+u(x)=A+V.$$

(2). Prove that if $$A$$ is a constant and $$\lim_{x\to y}u(x)=V$$ then $$\lim_{x\to y}Au(x)=AV.$$

(3). Prove that if $$\lim_{x\to y}u_1(x)=V_1$$ and $$\lim_{x\to y}u_2(x) =V_2$$ then $$\lim_{x\to y}u_1(x)+u_2(x)=V_1+V_2.$$

(4). Prove that if $$\lim_{x\to y}u_1(x)=0=\lim_{x\to y}u_2(x)$$ then $$\lim_{x\to y}u_1(x)u_2(x)=0$$.... See (**) below.

Suppose $$\lim_{x\to y}f(x)=L$$ and $$\lim_{x\to y}g(x)=M.$$ Let $$f(x)=L+h(x)$$ and $$g(x)=M+i(x).$$

By (1) with $$A=-L$$ and $$u=f$$ and $$V=L$$ we have $$\lim_{x\to y}h(x)=0.$$ Similarly we have $$\lim_{x\to y}i(x)=0.$$

Now $$f(x)g(x)-LM=h(x)M+i(x)L+h(x)i(x).$$ By (2) we have $$\lim_{x\to y}h(x)M=0\cdot M=0.$$ And similarly $$\lim_{x\to y}i(x)L=0.$$ So by (3) with $$u_1(x)=h(x)M$$ and $$u_2(x)=i(x)L$$ and $$V_1=V_2=0,$$ we have $$\lim_{x\to y}h(x)M+i(x)L=0.$$

By (4) we have $$\lim_{x\to y}h(x)i(x)=0.$$ So by (3) with $$u_1(x)=h(x)M+i(x)L$$ and $$u_2(x)=h(x)i(x)$$ we have $$\lim_{x\to y}f(x)g(x)-LM=0.$$

Finally by (1) with $$u(x)=f(x)g(x)-LM$$ and $$A=LM$$ we have $$\lim_{x\to y}f(x)g(x)=LM.$$

(**). To prove (4): Given $$e>0$$ let $$e'=\min (1,e).$$ Note that $$0<(e')^2\le e'\le e.$$ For $$j\in \{0,1\}$$ take $$d_j>0$$ such that $$0<|x-y| Let $$d=\min(d_1,d_2).$$ Then $$d>0$$ and $$0<|x-y|

• Good proof +1. However does my proof still make sense.
– user716881
Aug 2 '20 at 17:44