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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?

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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 $0<|x-y|<d_j\implies |u_j(x)|<e'.$ Let $d=\min(d_1,d_2).$ Then $d>0$ and $0<|x-y|<d\implies |u_1(x)u_2(x)|<(e')^2\le e.$

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

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