By definition, a pseudo-effective $\mathbb{R}$-divisor is the limit of effective $\mathbb{R}$-divisors in $N^1(X)$, I was wondering what is the example of pseudo-effective divisor which is not an effective divisor?
One thing I don't understand is why the limit of effective divisor may not be effective.
The simplest example I can think of is this:
Let $X$ be a $K3$ surface with $\rho(X)=3$ and with no $(-2)$-curves. (Surfaces like this really do exist.)
Then the pseudoeffective cone $\operatorname{PsEff}(X)$ (which in this case is equal to the nef cone) will be a standard "round" cone in $\mathbf R^3$. (That is, by changing basis of $\operatorname{Pic}(X) \otimes \mathbf R$ we could transform it to the cone $a^2-b^2-c^2 \geq 0$.)
Every boundary ray of this cone is extremal, but only countably many of them contain a rational (or integral) point. So any of the uncountably many other rays cannot be in the effective cone.
A more subtle problem (and maybe what the OP really wants) is to find an example where there is an integral divisor class $D$ such that $D \in \operatorname{PsEff}(X)$ but $D$ is not effective. For this, one can consult Example 1.5.1 in Lazarsfeld, Positivity in Algebraic Geometry, Volume 1.
In brief, this class of examples is as follows: $X$ is a ruled surface of the form $\mathbf P_C (\mathcal E)$ where $C$ is any curve of genus at least 2, $\mathcal E$ is a "sufficiently general" semistable bundle of degree 0 on $C$, and $D=c_1(O_X(1))$.
In this setup, $D$ spans an extremal ray of $\operatorname{PsEff}(X)$, but one can show that there is no effective curve $C$ with $[C] \in \mathbf R_+ \,D$.