If $f$ is plurisubharmonic (not identically $-\infty$) on a neighbourhood of $0$ then the [Lelong number](https://en.wikipedia.org/wiki/Lelong_number) of $f$ at $0$ is defined by $$\nu\_{f}(0) = \liminf\_{|z|\rightarrow 0}\dfrac{f(z)}{\log|z|}.$$


My question: How about $\displaystyle\limsup\_{|z|\rightarrow 0}\dfrac{f(z)}{\log|z|}$? Does it represent somesort of singularity at $0$ in literature? Can we construct a plurisubharmonic function $f$ such that $$1 = \displaystyle\liminf\_{|z|\rightarrow 0}\dfrac{f(z)}{\log|z|} < \displaystyle\limsup\_{|z|\rightarrow 0}\dfrac{f(z)}{\log|z|}<\infty?$$



A question about Lelong number
Your first question is unclear. The answer to the second question is positive: you can have a plurisubharmonic function such that $f(x\_k)=-\infty$
on some sequence $x\_k\to 0$. For this function $\limsup\_{x\to 0}f(x)/\log|x|=+\infty$. Then by small modification you can make this $\limsup$ finite but as large as you wish.


The simplest example is constructed as follows: in dimension $2$, take
$$f(z)=\max\left\{\sum\_0^\infty a\_n\log|1-z/z\_n|,M\log|z|\right\}$$
where $M>0,z\_n\to 0, a\_n>0$ and $\sum a\_n|\log|z\_n||<\infty.$ Then
the upper limit of the expression defining Lelong's number is $M$, while
the Lelong number itself is independent of $M$.


Same example works in higher dimension: just consider a function which depends only on $2$ variables.


