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I'm trying to give a proof for the following statement:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$. If $\limsup\limits_{x\rightarrow x_0} f(x)\leq \beta$, then for all $\epsilon>0$ there exists $\delta>0$ such that $|x-x_0|<\delta$ implies $f(x)\leq \beta + \epsilon$

I have this $\limsup$ definition:

$\limsup\limits_{x\rightarrow x_0} f(x)= \sup\{ z\space |\space \exists \{x_n\}_\mathbb{N} \rightarrow x_0 \text{ s.t. } \{f(x_n)\}_\mathbb{N}\rightarrow z\} \in\mathbb{R}\cup\{\pm\infty\}$

HELP!

TYPO:

To avoid things like “$\pm\infty$ plus other thing”, suppose that the limit actually exists, i.e., $\limsup\limits_{x\rightarrow x_0} f(x) \in\mathbb{R}$

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  • $\begingroup$ From what I can see, since $\limsup_{x \to x_0} f \leq \beta $ means on $|x - x_0| < \delta$, $\sup f(x) < \epsilon + \beta$. So certainly $f \leq \sup < \epsilon + \beta$ $\endgroup$ – IAmNoOne May 22 '18 at 8:36
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With the definition of $limsup$ you have this statement is false. The set in the definition can be empty in which case the supremum has to be taken as $-\infty$ (according to standard convention). Let $f(x)=\frac 1 {|x|}$ if $ x \neq 0$ and 0 if $x=0$. According to your definition $\limsup_{x \to x_0} f(x) =-\infty$. So the hypothesis holds with $\beta =0$. But there is no $\delta$ such that $f(x)\leq 0+\epsilon$ for $|x| <\delta$. The answer changes if your function $f$ is bounded. In that case the result claimed is true. I will provide a proof if you modify teh question accordingly.

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  • $\begingroup$ Actually the statement doesn’t mention anything about the boundness of a function. But, if it works, we can suppose that the limit actually exists and it’s a real number. $\endgroup$ – Sergio Ivan May 22 '18 at 13:05

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