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I know that if $\limsup_{x\to\infty}f(x)<L$ then there exist $x_{0}$ such that $f(x)<L$ for all $x\geq x_{o}$.

Is the following true:

If $\limsup_{x\to\infty}f(x)<L$ then, for any $\epsilon>0$, there exist $x_{0}$ such that $f(x)<L+\epsilon$ for all $x\geq x_{o}$.

(If the above is true, does $x_{o}$ depends on $\epsilon$?) Thanks

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    Did you mean this, very similar property? If $\limsup_{x\to\infty} f(x)\le L$ then there is an $x_0$ such that $f(x) for $x>x_0$. (The same is, of course, true if $\limsup_{x\to\infty} f(x)=L$; it is part of one of equivalent definitions of limit superior that appear in literature.) This is almost the same as you wrote, but $<$ is replaced by $\le$ or $=$.2012-04-17
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    Yes, this is exactly what I want. DO you have a reference for this theorem!2012-04-17
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    In many books this is part of definition of limsup. Could you edit the question and put your definition of limsup there? Or at least the name of the book you're studying or link to the lecture notes you're studying?2012-04-17

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