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In this answer, the limit is factored and then computed as the product of the limits of the factors. Can we always do this with limits? If not, under what circumstances can it be done?

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If $f$ and $g$ stay in a small neighborhood, then $f\cdot g$ also stays in a small neighborhood, as

$$|f(x)-F|<\delta_f\land|g(x)-G|<\delta_g\implies\exists\,\delta_{fg}:|f(x)\cdot g(x)-F\cdot G|<\delta_{fg}.$$

This is justified by $$|fg-FG|=|(f-F)(g-G)+G(f-F)+F(g-G)|<\delta_f\delta_g+|G|\delta_f+|F|\delta_g.$$

So if $f$ and $g$ have limits $F$ and $G$, so has $f\cdot g$, and it is $F\cdot G$.

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When the limit of the factors exists.

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  • $\begingroup$ Only the limit of the factors need to exist. This will in turn make sure that the limit of the product exist, and that the limit is the product of the limits. $\endgroup$ – Martin Apr 30 '15 at 8:01
  • $\begingroup$ Yeah you're right. I will edit it. $\endgroup$ – ThorbenK Apr 30 '15 at 8:06
  • $\begingroup$ Yves got the answer for greater detail. Had I more rep, you would have gotten an upvote for brevity, speed and clarity. $\endgroup$ – Fredrik P Apr 30 '15 at 8:52
  • $\begingroup$ @FredrikP: I made the upvote for you. $\endgroup$ – Yves Daoust Apr 30 '15 at 9:31
  • $\begingroup$ @YvesDaoust: Thanks! $\endgroup$ – Fredrik P Apr 30 '15 at 9:36

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