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This table has a column for "formal definitions" and another for "limit definitions" of the family of Bachmann-Landau notations.

What exactly are limt definitions and how do they relate to the formal definitions?

Since the notation is used to describe the asymptotic growth of functions, I thought the formal definitions were "limit" definitions themselves.

I'm especially not sure how the limit definitions for O, Ω and Θ are equivalent to their respective formal definitions. In the book Introduction to Algorithms, only the limit definition of Θ is given and it appears as a theorem (see Theorem 3.1)

user51462
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  • Is there a reason why you think they are not equivalent? – angryavian Jul 16 '22 at 01:22
  • I don't doubt they are equivalent, I'm just not sure how. A comment on another post seems to say that they aren't definitions, which confuses me. – user51462 Jul 16 '22 at 01:37
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    For big-O, notice that Wikipedia's limit definition uses $\limsup$, while the other post uses $\lim$. As the commenter points out, with $\lim$ there is only a one-way implication instead of an equivalence. // To focus your question, I would pick one row of the table and try to prove (or disprove) whether the two definitions are equivalent. – angryavian Jul 16 '22 at 03:16
  • Thank you @angryavian, that clears it up. Could you please explain why lim only has a one-way implication instead of equivalence? – user51462 Jul 16 '22 at 23:46
  • The limit definitions given here and elsewhere are confusingly all in terms of lim instead of lim sup and lim inf. – user51462 Jul 16 '22 at 23:52
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    For your first question read Did's comment again. If $f(n) = \begin{cases} n & \text{$n$ even} \ 0 & \text{$n$ odd}\end{cases}$ and $g(n)=n$, then $f(n)=O(g(n))$ but $f(n)/g(n)$ doesn't have a limiit. – angryavian Jul 17 '22 at 00:23
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    For your new link, note that #1 and #2 are talking about little-$o$ and little-$\omega$, for which the "limit definitions" use $\lim$ and agree with Wikipedia. Their #3 is incorrect though; $f(n)/g(n) \to c$ implies $f(n) \in \Omega(g(n))$, but the converse isn't necessarily true since the limit of $f(n)/g(n)$ might not exist. – angryavian Jul 17 '22 at 00:25
  • Thanks so much @angryavian. – user51462 Jul 17 '22 at 07:26

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