The relationship between LCTVS and projective limit of a projective family of norm spaces. To begin with, the projective limit of a projective family of norm spaces is a LCTVS (locally convex topological space).
However, on the other hand, I want to find an example of LCTVS such that it cannot be the projective limit of any projective family of norm spaces. I tried to work on the generating seminorms but I got nothing.
To sum up, my questions are:

*

*an example(above-mentioned)

*LCTVS under what extra conditions could be isomorphic to a projective limit of a projective family of norm spaces?

 A: Let $E$ be a Hausdorff locally convex space and $\{p_\alpha: \alpha \in I\}$ be a directed family of seminorms that defines the topology of E (f.e. it can be the set of all continuous seminorms on $E$). Let $E_\alpha = E/p_\alpha^{-1}(\{0\})$ for all $\alpha \in I$. $E_\alpha$ is a normed space wrt to norm that is induced by $p_\alpha$ (i.e. $||\pi_\alpha(x)||_\alpha = p_\alpha(x)$, where $\pi_\alpha:E \rightarrow E_\alpha$ is the canonical projection and $x \in E$). If $p_\alpha \ge c p_\beta$, where $c > 0$, then there is a canonical map $E_\alpha \rightarrow E_\beta$ that is continuous wrt to the foregoing norms. Thus, spaces $E_\alpha$ form a projective family of normed spaces. It can be easily shown that $E$ is canonically isomorphic to the projective limit of this projective family (note that if $E$ is not necessarily Hausdorff, then this projective limit is canonically isomorphic to $E/\overline{\{0\}}$).
Thus, every Hausdorff locally convex space is isomorphic to a projective limit of normed spaces. Since the projective limit of a family of Hausdorff locally convex spaces is again Hausdorff, it follows that the Hausdorff condition is also necessary. (For non-Hausdorff spaces you can define $E_\alpha$ as a seminormed space that is algebraically equal to $E$ with seminorm $p_\alpha$. Thus, arbitrary locally convex space is a projective limit of a family of seminormed spaces.)
There is a notable addition to the foregoing construction. We can also consider the completions of $E_\alpha$ and the continuations of the canonical maps $E_\alpha \rightarrow E_\beta$. The projective limit of this directed family is the completion of $E$. Thus, a Hausdorff locally convex space $E$ is a projective limit of a projective family of Banach spaces iff $E$ is complete (since projective limit preserves completeness).
A: Not every Hausdorff locally convex space is the projective limit of normed spaces:
To prove this, we associate to a locally convex space $(X,\tau)$ the sk-topology $\tau^{sk}$ as the group topology  on $X$ having $\{$ker$(p): p$ continuous seminorm on $X$} as a basis of the neutral element $0$ (if $X$ has a continuous norm, this is the discrete topology). $X$ is called sk-complete if the topological group $(X,\tau^{sk})$ is complete. It is easily seen that $\tau^{sk}$ "commutes" with products and subspaces and that products and closed subspaces of sk-complete spaces are sk-complete. Since normed spaces are (trivially) sk-complete, this shows that projective limits of normed spaces are sk-complete. Vice versa, the standard construction as in Matsmir's answer (factoring out the kernel $K_p$ of a seminorm $p$ to obtain a normed spaces $X_p=(X,p)/K_p$) yields a canonical map $I$ from $X$ into the projective limit of the spaces $X_p$ which, for Hausdorff spaces $X$, gives a topological embedding with dense range.
Now the crucial observation is that this is also an sk-$embedding$ (by the stability properties of the associated sk-topology) whose range is sk-dense (because the quotient maps $X\to X_p$ are surjective). If thus $X$ is sk-complete we obtain that $I$ is an isomorphism. We have thus proved
Theorem. A locally convex topological vector space is a projective limit of normed spaces if and only if it is Hausdorff and sk-complete.
Let us finally give a simple example:
Let $X=C([0,1])$ be the space of real-valued continuous functions endowed with the topology $\tau$ of pointwise convergence given by the seminorms $p_E(f)=\max\{|f(x)|:x\in E\}$ with $E \subseteq [0,1]$ finite. For an arbitrary function $g:[0,1]\to\mathbb R$ and every finite set $E$ there is a continuous function $f$ such that $g|_E=f|_E$ (e.g., by linear interpolation between the finitely many given values).
For the space $Y=\mathbb R^{[0,1]}$ of all functions (just a big cartesian product) endowed with the topology $\sigma$ of pointwise convergence (the product topology), this shows that the inclusion $(X,\tau^{sk})\to (Y,\sigma^{sk})$ is a hoemeomorphism with dense range. Thus, $X$ is not sk-complete (because complete subgroups of a Hausdorff topological group are closed).

I do not know whether this can be found in the literature and I would be grateful for a reference. The arguments are, of course, quite routine and similar to the answer of user131781 to this MO question https://mathoverflow.net/q/331286.
I have used (sequential) sk-completeness for other fancy things in my Springer Lecture Notes Derived Functors in Functional Analysis. The associated sk-topology has also been used by D. Vogt in Splitting of exact sequences of Fréchet spaces in the absence of continuous norms. J. Math. Anal. Appl. 297 (2004), 812–832.
