Consider $F$ either left or right exact functor from some nice abelian category to another nice abelian category. Then $L_\star(F)$ or $R^\star(F)$ are universal $\delta$ functors.
$\textbf{Q:}$ Consider inverse limit functor $\lim$ which is left exact and one has to assume AB4 axiom for abelian category. $R^i\lim=0$ for $i\neq 0,1$. Is it possible to have the following situation for universal $\delta$ functor? Let $L_\star(F)=\{F,L_1F,L_2F=0,L_3F,\dots\}$ with $L_3F\neq 0$. It seems from the proof of $R^i\lim=\{\lim,\lim^1,0,\dots\}$ forming universal $\delta$ functor, I would deduce $L_iF=0$ for $i\geq 3$ as I always have a unique $0$ map. In other words, once the functors $L_\star(F)$ vanishes at some position say $i$, then $L_j(F)=0$ for $j>i$.
Your deduction is correct; it follows from the universality of universal $\delta$-functors that derived functors are unique. See Weibel's An introduction to homological algebra, page 32 for the statement:
Thus it is enough to show that, if one knows that $L_{i}F=0$ for some $i$, then the $\delta$-functor $T=\{F,L_{1}F,\ldots,L_{i-1}F,0,0,0,0,\ldots\}$ is also a universal $\delta$-functor.
Indeed, first note that $T$ is a $\delta$-functor; given short exact sequences $\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\big\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\ua}[1]{\uparrow\raise.5ex\rlap{\scriptstyle#1}}$ \begin{array}{c} 0&\ra{}&A&\ra{}&B&\ra{}&C&\ra{}&0\\ &&\da{}&&\da{}&&\da{}&&\\ 0&\ra{}&A'&\ra{}&B'&\ra{}&C'&\ra{}&0 \end{array} we must show we get a commutative ladder of long exact sequences: \begin{array}{c} \cdots&\ra{}&T_{n-1}C&\ra{\delta}&T_{n}A&\ra{}&T_{n}B&\ra{}&T_{n}C&\ra{\delta}&T_{n+1}A&\ra{}&\cdots\\ &&\da{}&&\da{}&&\da{}&&\da{}&&\da{}\\ \cdots&\ra{}&T_{n-1}C'&\ra{\delta}&T_{n}A'&\ra{}&T_{n}B'&\ra{}&T_{n}C'&\ra{\delta}&T_{n+1}A'&\ra{}&\cdots\\ \end{array} Indeed we do, since for $n\leq i-1$, the commutative ladder is given by the very same for $L_{*}F$ (not assuming any terms are $0$), and for $n\geq i$, the following is indeed a commutative ladder of long exact sequences: \begin{array}{c} \cdots&\ra{}&L_{i-1}FC&\ra{\delta}&0&\ra{}&0&\ra{}&0&\ra{\delta}&0&\ra{}&\cdots\\ &&\da{}&&\da{}&&\da{}&&\da{}&&\da{}\\ \cdots&\ra{}&L_{i-1}FC'&\ra{\delta}&0&\ra{}&0&\ra{}&0&\ra{\delta}&0&\ra{}&\cdots\\ \end{array} with $0$s forever. Thus $T$ is a $\delta$-functor.
Now we just need to show that $T$ is universal. A (homological, covariant) $\delta$-functor $T$ is universal if, given any other $\delta$-functor $S$ and a natural transformation $f_{0}:S_{0}\to T_{0}$, there exists a unique morphism $\{f_{n}:S_{n}\to T_{n}\}$ of $\delta$-functors that extends $f_{0}$. Note first that since $L_{*}F$ is universal, if there exists a $\delta$-functor $S$ and a $f_{0}:S_{0}\to L_{0}F$, then there exists a unique $\{f_{n}:S_{n}\to L_{n}F\}$ extending $f_{0}$. What is this $\{f_{n}\}$? See that for $n=i-1$ and a short exact sequence $0\to A\to B\to C\to0$, we must have \begin{array}{c} \cdots&\ra{}&S_{i-1}C&\ra{\delta}&S_{i}A&\ra{}&S_{i}B&\ra{}&S_{i}C&\ra{\delta}&S_{i+1}A&\ra{}&\cdots\\ &&\da{f_{i-1}}&&\da{f_{i}}&&\da{f_{i}}&&\da{f_{i}}&&\da{f_{i+1}}\\ \cdots&\ra{}&L_{i-1}FC&\ra{\delta}&0&\ra{}&0&\ra{}&0&\ra{\delta}&L_{i+1}FA&\ra{}&\cdots\\ \end{array} Thus $f_{i}=0$. For $f_{i+1}$, see that it too is forced to be $0$, since \begin{array}{c} S_{i}C&\ra{\delta}&S_{i+1}A\\ \da{0}&&\da{f_{i+1}}\\ 0&\ra{\delta}&L_{i+1}FA \end{array} must commute and $A$ can be arbitrary. Via induction, $f_{j}=0$ for $j\geq i$.
Now, simply observe that the very same $\{f_{n}\}$ is a morphism of $\delta$-functors $S\to T$ that extends $f_{0}$. Is it unique, thus proving universality of $T$? Yes, it must be! As $T_{j}=0$ for $j\geq i$, there are no other possible maps $f_{j}:S_{j}\to T_{j}$ other than the zero map.
Therefore $T=\{F,L_{1}F,\ldots,L_{i-1}F,0,0,0,\ldots\}$ is a universal $\delta$-functor with $T_{0}=F$, as is $L_{*}F$, and so by Weibel, the two are isomorphic.