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Abbott's Understanding Analysis, while proving 'If $\lim (a_n)\rightarrow a$ and $\lim (b_n)\rightarrow b$ then $\lim (a_nb_n)\rightarrow ab$ (where both the sequences and their limits are in $\Bbb R$)', leaves the case of $a=0$ to exercise, and for case $a\ne 0$, goes thus:

Let $\epsilon>0$ be arbitrary. If $a\ne 0$, then we can choose $N_1$ so that $n\ge N_1$ implies $|b_n-b|<\frac1{|a|}\frac{\epsilon}2$. Since $b_n$ is convergent, there exists $M>0$, such that $|b_n|\le M$ for all $n\in N$. Now we can choose $N_2$ so that $|a_n-a|<\frac1M\frac{\epsilon}2$ whenever $n\ge N_2$. Now pick $N=\max \{N_1,N_2\}$ and observe that if $n\ge N$ then $$|a_nb_n-ab|\le |b_n||a_n-a|+|a||b_n-b|\le M|a_n-a|+|a||b_n-b|<M\frac {\epsilon}{M2}+|a|\frac {\epsilon}{|a|2}=\epsilon.$$

My question is for case $a=0$:

Instead of suggesting to go this way: $$|a_nb_n-ab|\le |b_n||a_n-a|+|a||b_n-b|=|b_n||a_n-a|\le M|a_n-a|<M\frac {\epsilon}{M2}<\epsilon,$$ the book gives a three part exercise for this case: enter image description here enter image description here

Why the 'easy' way is (perhaps) wrong way?

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  • $\begingroup$ What is the "Algebraic Limit Theorem"? Most of the basic results about limits don't have commonly-used names. $\endgroup$ – DanielWainfleet Jan 10 '17 at 7:15
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What you've written in your suggestion essentially has the solution to exercise 2.3.7.(a) in it. If $b_n\rightarrow b$ and $a_n\rightarrow 0$ , then the proof that $a_nb_n \rightarrow 0$ is much easier because $|a_nb_n|=|a_n||b_n| \leq \frac{\epsilon}{2M}M<\epsilon$.

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Just want to show you how to prove that $a_n\to a$ and $b_n\to b$ implies $a_nb_n\to ab$ by handling all cases at once. No extra work for the case $a=0.$

Take the least (or any) $n_1$ such that $n\geq n_1\implies |a-a_n|<1.$ Take the least (or any) $n_2$ such that $n\geq n_2\implies |b-b_n|<1.$ Let $n_3=\max (n_1,n_2).$

Let $M=1+\max (|a|,|b|).$ Then $|b|<M$. And also $n\geq n_3\implies |a_n|<M.$

Given $\epsilon >0,$ take $n'\geq n_3$ such that $n\geq n'\implies |a-a_n|<\epsilon /2M.$ Take $n''\geq n_3$ such that $n\geq n''\implies |b-b_n|<\epsilon /2M.$ Let $n'''=\max (n',n'').$

Now $$n\geq n'''\implies |ab-a_nb_n|=|(a-a_n)b+a_n(b-b_n)|\leq$$ $$\leq |a-a_n|\cdot |b|+|a_n|\cdot |b-b_n|\leq |a-a_n|M+M|b-b_n|<\epsilon.$$

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  • $\begingroup$ Thank you for providing this insight! $\endgroup$ – Silent Jan 10 '17 at 9:51
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    $\begingroup$ The difficulty comes in the last line, where we need a common upper bound ($M$) for $|b|$ and for $|a_n|$ that does not depend on $ n.$ We could shorten the proof by first proving that a convergent sequence is a bounded sequence, and then we don't need $n_1,n_2,n_3$ : Just let $M=\max (|b|, \max \{|a_n|:n\in \mathbb N\}).$,,,, We can also use the method in the A to show that if functions $f(x), g(x)$ are continuous at $x_0$ then $f(x)g(x)$ is continuous at $x_0$. $\endgroup$ – DanielWainfleet Jan 10 '17 at 10:27

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