By definition of limit superior, we have $\limsup_{n\to \infty}x_n=\lim_{n\to \infty}(\sup_{m\geq n}x_m)$. Whats is the definition of $\limsup_{x\to 0^+}f(x)$?
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1If you had to guess, what would you guess? – Arthur Nov 21 '22 at 13:31
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@Arthur For every $x>0$, there exists $n\in \mathbb{N}$ such that $\frac{1}{n}<x$. So $x\to 0^+$ implies that $n\to +\infty$. Hence $\limsup_{x\to x^+}f(x)$ turns into $\limsup_{n\to +\infty}f(\frac{1}{n})$. Am I correct? – M.Ramana Nov 21 '22 at 13:43
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No. See: https://en.wikipedia.org/wiki/Limit_inferior_and_limit_superior – Fred Nov 21 '22 at 13:44
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@Fred Thanks for comment. Could you please tell me what the defnition of $\limsup_{x\to a}f(x)$ is? – M.Ramana Nov 21 '22 at 13:51
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It's the limit of the maximum values so far of $f(x)$ as $x\to 0^+.$ – Adam Rubinson Nov 21 '22 at 13:55
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Your idea has some merits, but it misses a few important subtleties. More crucially, though, is that it is nowhere near the actual definition. The usual definition looks (visually) almost identical to your $n\to\infty$ definition, only with small but necessary changes here and there. (And also it never cares about integers.) Try again, see if you can make a definition that works. – Arthur Nov 21 '22 at 13:55
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@Arthur Thank you so much for your explanation. Sure, I'll try again. – M.Ramana Nov 21 '22 at 14:28
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@Arthur I've tried again and agian but I couldn't make a definition that works. Could you please help me about it? I wrote a question here: https://math.stackexchange.com/questions/4581522/explain-about-these-limit-superior-and-limit-inferior – M.Ramana Nov 23 '22 at 03:40
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@Arthur But no one helped me there. So I asked the definition of $\limsup_{x\to 0^+}$ here. But I have no idea about it. Could you please help me? – M.Ramana Nov 23 '22 at 03:42
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The definition of $\limsup_{x\to0^+}f(x)$ is $$ \lim_{x\to 0^+}\left(\sup_{0<y<x}f(y)\right) $$ Can you see how similar this is to your $\limsup_{n\to\infty}x_n$ definition?
Arthur
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