I think I understand the limit by definition to some extent, but why $ε/2$? Why not $ε/3$ or $ε/4$? I fail to grasp the utility of dividing epsilon and I often encounter it in problems of limits and continuity. Any explanation is appreciated.
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1There is no difference: you can also put $ε/3$ or $ε/4$. Traditionally one puts either $\varepsilon/2$ or $\varepsilon$ small as desired. – Sebastiano Jan 22 '23 at 11:59
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6It basically depends on what you're trying to prove, and generally the choice of what coefficient to use (or what power to put $\varepsilon$ at if powers are involved, but that's less frequent) is just to make the proof more "elegant" by the time you finish it. – Bruno B Jan 22 '23 at 12:02
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@Sebastiano thank you for your reply. But, is that a rule? That if a limit is equal to zero then |f(x)|≤ε. In other words, is this applicable if the limit is equal to some other real number? What about infinity? – 真個I Jan 22 '23 at 12:03
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@BrunoB Is this applicable if limit f(x)=a, where a is different from zero? Thank you for your reply! – 真個I Jan 22 '23 at 12:05
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2In that case you'll just go with $|f(x) - a| \leq \lambda\varepsilon$, with whatever $\lambda$ you wish to use. – Bruno B Jan 22 '23 at 12:08
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7When I was at university, I did not understand the reason because I sometimes had to put $\varepsilon/2$ when proving the limit theorem of a sum. All because when you make the addition required by the theorem the total sum is $\varepsilon/2+\varepsilon/2=\varepsilon$. It is just an elegance to have $\varepsilon$. You could also have in this theorem a total sum of $2\varepsilon$ which is the same as having $\varepsilon$. – Sebastiano Jan 22 '23 at 12:08
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@BrunoB I am agree with you totally. – Sebastiano Jan 22 '23 at 12:09
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In this answer I give about 12 equivalent conditions for limits. Which one happens to be more convenient for the task at hand varies from individual to individual and task to task. – peek-a-boo Jan 22 '23 at 19:11