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Let $f: [0, \infty) \rightarrow \mathbb R$. What is the definition of $$ \limsup_{x \rightarrow a} f(x)$$ for $a \in [0, \infty)$?

Dominic Michaelis
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user61184
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    See here. – Stefan Hansen Feb 15 '13 at 08:53
  • Given that all the definitions offered (Wikipedia link, answer below) put the focus on neighborhood formulations, I thought I would mention that I find it much easier to intuitively grasp and work with the notion $\limsup_{x \rightarrow a} f(x)$ by thinking of it as the largest of the functional values you can get when using sequences that approach $a$ (where none of the terms of the sequence can be equal to $a$). Of course, one would first say "supremum of the" and then prove that the supremum can always be achived for some sequence. – Dave L. Renfro Feb 15 '13 at 15:15

1 Answers1

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This definition is given in some books: $$\limsup_{x \rightarrow a} f(x)=\inf_{0<r}\sup_{x\in ((a-r)\cup(a+r))\cap[0,\infty)}f(x)$$

or simply: $$\limsup_{x \rightarrow a} f(x)=\inf_{0<\epsilon}\space\space\sup_{0<|x-a|<\epsilon}f(x)$$

This is the deleted upper limit. Undeleted one is defined by:

$$Limsup_{x \rightarrow a} f(x)=\inf_{0<\epsilon}\space\space\sup_{|x-a|<\epsilon}f(x)$$