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"Definition: For every $\epsilon > 0$ there is some $\delta > 0$ such that, for all $x$, if $0 < |x - a| < \delta$, then $|f(x) - l| < \epsilon$ This is if $f$ approaches the limit $l$ near $a$. "

It says for some $\delta$ so all we need to do is prove there is only one $\delta$ such that $|f(x) - l| < \epsilon$ which could mean $\delta = \epsilon$ or we could just choose one delta right?

Thanks!

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  • $\begingroup$ "There is only"? There is at least one. $\endgroup$ – Git Gud Oct 10 '14 at 13:57
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    $\begingroup$ I'm not sure I understand your question correctly, but if you're asking if it suffices to prove there exists at least one $\delta$ for any given $\varepsilon$, yes, that is enough. Even once you show the existence of one, you'll have infinitely many $\delta$ that suffice since you could take $\delta/2$, $\delta/3$, and so on. Then use $\cdots<\delta/3<\delta/2<\delta$. $\endgroup$ – Clayton Oct 10 '14 at 13:58
  • $\begingroup$ So you have to prove there is AT LEAST one $\delta$? And that proves that the limit is $l$? Really? $\endgroup$ – Amad27 Oct 10 '14 at 13:58
  • $\begingroup$ Not only one, if say, works for $\delta_1>0$ so certainly works for any $0<\delta<\delta_1$ $\endgroup$ – Jose Antonio Oct 10 '14 at 13:59
  • $\begingroup$ Yes. But you have to prove it works for AT LEAST one $\delta$? $\endgroup$ – Amad27 Oct 10 '14 at 13:59
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Semantically, you have the right idea. "...There is some $\delta > 0$" means that only one suffices to satisfy the definition.

But don't forget the first part of the statement: "For all $\epsilon > 0$..." It is very likely that the $\delta$ you supply depends on $\epsilon$. In some cases it may suffice to let $\delta = \epsilon$, but certainly not all.

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