The problem is

If $a_n$ converges to $a$ and $b_n$ converges to b, prove that $\displaystyle \lim_{n \rightarrow \infty} a_n b_n = ab$.

My book proves it like this

$|a_nb_n - ab| = |(a_nb_n -a_n b)+(a_n b - ab)| \leq |a_n||b_n - b| + |b||a_n-a|$

Since $a_n$ converges, there exists an $M >0$ such that $|a_n| \leq M$ for all $n$.


$|a_nb_n - ab|\leq M|b_n - b| + |b||a_n-a|$.

Let $\epsilon >0$. Since $a_n \rightarrow a$, there exists $n_1$ such that

$|a_n-a| < \frac{\epsilon}{2(|b|+1)}$

for all $n \geq n_1$.

Since $b_n \rightarrow b$, there exists $n_2$ such that

$|b_n-b| < \frac{\epsilon}{2M}$

for all $n \geq n_2$.

Let $n = \max(n_1, n_2)$, then

$|a_nb_n - ab|< M (\frac{\epsilon}{2M} )+ |b| ( \frac{\epsilon}{2(|b|+1)}) < \epsilon$

My question about this proof is why they chose $|a_n-a| < \frac{\epsilon}{2(|b|+1)}$ instead of $|a_n-a| < \frac{\epsilon}{2|b|}$

wouldn't that have been a nicer choice?

Thank you.


1 Answer 1


Substantially because $ b $ might be $0$, in which case that "bound" would not be a "bound".

  • $\begingroup$ So they chose their bound to prevent division by zero if b is zero? $\endgroup$
    – 1233211
    Dec 22, 2016 at 7:19
  • 1
    $\begingroup$ Most importantly, to prevent a $\le+\infty\cdot\epsilon $ from appearing on the RHS of their estimates: that would make them completely uninformative. $\endgroup$
    – user228113
    Dec 22, 2016 at 7:23

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