upper limit. Definition for function.

Question

reading the book, I came across definitions MCshane "Integration".

If $f$ is defined on a set $E$ and $x_{0}$ is an accumulation point of $E$ then $$\limsup\limits_{x\to x_{0}}f(x)=\inf\{M_{\delta}(f;x_{0}) : \delta>0\},$$ where $$ M_{\delta}(f;x_{0})=\sup f(N_{\delta}(x_{0})\cap E\setminus x_{0}\}$$ $$N_{\delta}(x_{0})=(x_{0}-\delta,x_{0}+\delta) $$ Its my first time meeting that kind defitnion of upper limit. Can anyone explain what this definition is about?

Are you familiar with the usual definition of the limit superior of a function/sequence? — G. Chiusole, Jun 21 '20 at 16:41
I tried to find different definitions but even wikipedia give a similar definition. :/ — Jacob12xx, Jun 21 '20 at 16:44

score 0 · Answer 1 · answered Jun 21 '20 at 16:46

If $M=\limsup_{x\to x_0}f(x)$, then $M$ is the only element of $\Bbb R\cup\{\pm\infty\}$ such that both of these conditions hold:

if $K>M$, then there is some interval $(x_0-\varepsilon,x_0+\varepsilon)$ centered at $x_0$ such that if $x\in(x_0-\varepsilon,x_0+\varepsilon)\cap E$, then $f(x)<K$;
if $K<M$, then for every interval $(x_0-\varepsilon,x_0+\varepsilon)$ centered at $x_0$ there are numbers $x\in(x_0-\varepsilon,x_0+\varepsilon)\cap E$ such that $f(x)>K$.

Using this, you can check that, for instance, $\limsup_{x\to0}\sin\left(\frac1x\right)=1$. Or that, if the limit $\lim_{x\to x_0}f(x)$ exists (in $\Bbb R\cup\{+\infty\}$), then $\limsup_{x\to x_0}f(x)=\lim_{x\to x_0}f(x)$.

In the first case is true for all x, and second case shouldn't be exist K? — Jacob12xx, Jun 21 '20 at 22:18
In the first case, yes, what I wrote was that, for all $x\in(x_0-\varepsilon,x_0+\varepsilon)\cap E$, $f(x)<K$. And, no, I did not mean that there is some $K$. I meant for all $K$. — José Carlos Santos, Jun 21 '20 at 22:32

upper limit. Definition for function.

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