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