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?

Answer 1

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 $\limsup_{x\to x_0}f(x)=\lim_{x\to x_0}f(x)$.