1. Introduction
In this paper, all graphs considered are simple and undirected. Let 
$G=(V(G),E(G))$ be a graph with order 
$n=|V(G)|$ and size 
$m = |E(G)|$. For a subset 
$W\subseteq V(G)$, we use 
$G[W]$ to denote the subgraph of 
$G$ induced by 
$W$, and use 
$G-W$ to denote the induced subgraph 
$G[V(G)\backslash V(W)]$. An edge subset 
$M \subseteq E(G)$ is called a matching of 
$G$ if no two distinct edges in it share a common vertex, and let 
$\mathcal{M}(G)$ be the set of matchings (including the empty matching) of 
$G$. To study statistical physics, Heilmann and Lieb [Reference Heilmann and Lieb8] gave a formal definition of matching polynomial of 
$G$ as follows:
\begin{equation*}\mathscr{M}(G,x)=\sum_{M\in \mathcal{M}(G)}(-1)^{|M|}x^{|V(G)\backslash V(M)|}.\end{equation*}As is known, the matching polynomial is one of the classical graph polynomials in algebraic combinatorics and has a wide range of applications. In particular, it was used to construct infinitely many bipartite Ramanujan graphs [Reference Marcus, Spielman and Srivastava13]. For more details, we refer readers to the related chapters of the monographs [Reference Godsil5, Reference Shi, Dehmer, Li and Gutman16].
In 2020, Mohammadian [Reference Mohammadian14] introduced the Laplacian matching polynomial of 
$G$ defined as follows:
\begin{align*}\mathscr{L}(G,x)=\sum_{M\in \mathcal{M}(G)}(-1)^{|M|}\prod_{v\in V(G)\backslash V(M)}\left(x-d_{G}(v)\right),\end{align*} where 
$d_{G}(v)$ is the degree of 
$v \in V(G)$ and 
$V(M)$ is the set of vertices incident to some edge in 
$M$. As pointed out in [Reference Li and Yan12], the Laplacian matching polynomial of a graph is identical with its average Laplacian polynomial introduced by Zhang and Chen [Reference Zhang and Chen21], who were inspired by the celebrated result of Godsil and Gutman [Reference Godsil and Gutman6] that the matching polynomial is just the average of adjacency characteristic polynomials of all signed graphs with underlying graph 
$G$. Nowadays, the Laplacian matching polynomial of 
$G$ has been attracting much more attention with its fantastic properties. Mohammadian [Reference Mohammadian14] proved that all roots of this polynomial are nonnegative and real. Zhang and Chen [Reference Zhang and Chen21] revealed the relations between this polynomial and graph structures. Wan, Wang and Mohammadian [Reference Wan, Wang and Mohammadian17] investigated the location of its zeros. Li and Yan [Reference Li and Yan12] gave a combinatorial explanation of its coefficients. See [Reference Wan, Wang and Wang18, Reference Wang, Cui and Cioaba20] for more new results. For convenience, the Laplacian matching roots of 
$G$ are those of its Laplacian matching polynomial, where the largest one is called the Laplacian matching index of 
$G$.
The spectral radius 
$\rho(G)$ of a graph 
$G$ is that of its graph matrix. A real number 
$\gamma$ is called a limit point of spectral radii of graphs if there exists a sequence of graphs 
$\{G_k\in\mathcal{G}\, |\, k\in \mathbb{N}\}$ such that
\begin{align*}
\rho(G_i) \neq \rho(G_j) \,\text{whenever i} \neq \text{j}, \text{and} \lim_{k \rightarrow \infty}\rho(G_k)=\gamma.
\end{align*} Historically, it originated from A.J. Hoffman [Reference Hoffman9] on the limit points of spectral radii (or say, the largest roots of characteristic polynomials) of nonnegative symmetric integral matrices by means of adjacency matrices of graphs. Later on, the Hoffman program of graphs [Reference Belardo, Marzia, Simić and Wang2] (with respect to a fixed spectral radius) mainly determine all the limit points of the set 
$\{\rho(G)\vert G\in\mathcal{G}\}$ and characterise all 
$G$’s such that 
$\rho(G)$ does not exceed a fixed limit point. As an important application, Jiang and Polyanskii [Reference Jiang and Polyanskii11] used the Hoffman program to attack a longstanding problem on equiangular lines. So far, Hoffman’s program of graphs has been extensively studied in the setting of the Laplacian, the signless Laplacian, the signed adjacency, the Hermitian adjacency, the skew-adjacency and 
$A_\alpha$-matrices of graphs as well as the adjacency tensors of hypergraphs. See a very recent survey [Reference Wang, Wang, Maurizio, Belardo and Wang19] and a new paper [Reference Oliveira and Trevisan15] for more details.
In this paper, we will consider the Hoffman program for the Laplacian matching indices 
$\lambda(G)$ of graphs at the first attempt. More generally, the Laplacian matching index of 
$G$ is an algebraic invariant but not a spectral radius.
We introduce more notations. Let 
$G \cup H$ be the disjoint union of two graphs 
$G$ and 
$H$. We use 
$H \subseteq G$ to denote the subgraph 
$H$ of 
$G$, and use 
$H \subset G$ if it is proper. For a graph matrix 
$B$, the B-characteristic polynomial of 
$G$ is 
$\varphi_B(G,x)=\det(xI-B)$ with 
$I$ being the identity matrix. The spectral radius 
$\rho_B(G)$ of 
$G$ is the largest absolute value of all zeros of 
$\varphi_B(G,x)$. Let 
$A(G)=(a_{ij})$ be the adjacency matrix of 
$G$, where 
$a_{ij} = 1$ if 
$v_iv_j \in E(G)$ and otherwise 
$a_{ij} = 0$. The Laplacian matrix of 
$G$ is 
$L(G) = D(G)-A(G)$, where 
$D(G) = {\rm diag}(d_G(v_1), d_G(v_2), \cdots, d_G(v_n))$ is the diagonal matrix of vertex degrees. Let 
$\mu(G)$ and 
$\lambda(G)$ denote the largest root of 
$\mathscr{M}(G,x)$ and 
$\mathscr{L}(G,x)$, respectively. As usual, set 
$P_n$ and 
$C_n$ to be the path and cycle with order 
$n$, respectively. Let 
$F_{s,r}(s \gt 1,r\geq1)$ denote the tree obtained from 
$P_{s+r+1}$: 
$v_{0}v_{1}\cdots v_{s}u_{1}\cdots u_{r}$ and two isolated vertices 
$w_{1},w_{2}$ by adding edges 
$v_{1}w_{1}$ and 
$v_{s}w_{2}$.
The organisation of this paper is as follows: In Section 2, we introduce some basic lemmas and useful results. In Section 3, motivated by the Hoffman program w.r.t. Laplacian matrix (see the relation (1)), we determine all the limit points of the Laplacian matching indices of graphs less than 
$\tau = 2+\omega^{\frac{1}{2}}+\omega^{-\frac{1}{2}} (= 4.38+)$, where 
$\omega=\frac{1}{3}(\sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}}+1)$, and then characterise the connected graphs with the Laplacian matching indices less than 
$2+\sqrt{5}$. Finally, we put forward two problems for further study.
2. Preliminaries
As a starter, we give the relation between matching and Laplacian matching polynomials of a graph.
Lemma 2.1. ([Reference Zhang and Chen21])
Let 
$G$ be a graph. Then
\begin{equation*}\mathscr{M}(S(G),x)=x^{|E(G)|-|V(G)|}\mathscr{L}(G,x^{2}),\end{equation*} where 
$S(G)$ is the subdivision graph of 
$G$.
For a connected graph 
$G$ and 
$v\in V(G)$, Godsil defined the path-tree 
$T(G,v)$ of 
$G$ to be a tree which has as vertices the paths in 
$G$ that start at 
$v$ and where two such paths are adjacent if one is a maximal proper sub-path of the other, and he showed that 
$\mu(G)=\rho_{A}(T(G,v))$ [Reference Godsil4]. On the other side, Mohammadian [Reference Mohammadian14] proved that
\begin{equation}
\varphi_L(G,x) = \mathscr{L}(G,x)\;\; \text{iff~G~is~a~forest}.
\end{equation} So, 
$\rho_L(G) = \lambda(G)$. Along with the previous lemma, we immediately get the following corollary.
Corollary 2.2. Let 
$G$ be a connected graph and 
$S(G)$ be its subdivision with 
$v \in V(S(G))$. Then
\begin{equation*}
\lambda(G) = \mu^2(S(G)) = \rho_A^2(T(S(G),v)).
\end{equation*} Let 
$\Delta(G)$ and 
$\delta(G)$ be, respectively, the maximum and minimum degrees of 
$G$.
Lemma 2.3. Let 
$G$ be a connected graph with 
$H \subset G$ and 
$\ell(G)$ be the length of a longest path in 
$G$. Then
(i) [Reference Cvetković, Rowlinson and Simić3]
$\;\rho_{A}(H) \lt \rho_{A}(G)$.(ii) [Reference Wan, Wang and Mohammadian17]
$\lambda(H) \lt \lambda(G)$ and
\begin{equation*}\max\{\Delta(G)+1, \delta(G)+\sqrt{\Delta(G)}\} \leq \lambda(G) \leq \Delta(G)+2\sqrt{\Delta(G)-1}\cos{\frac{\pi}{2\ell(G)+2}}\end{equation*}with the left equality holds iff
$G$ is a star, and the right one holds iff
$G$ is a cycle.
Hoffman and Smith [Reference Hoffman, Smith and Fiedler10] defined an internal path of a graph 
$G$ as a walk 
$v_0,v_1,\cdots,v_k(k \geq 1)$ where the vertices 
$v_1,\cdots,v_k$ are distinct (
$v_0, v_k$ need not be distinct), 
$d_{G}(v_0) \gt 2$, 
$d_{G}(v_k) \gt 2$ and 
$d_{G}(v_i) = 2$ whenever 
$0 \lt i \lt k$. In terms of the definition, two types of internal paths (see Fig. 1) are introduced as follows:
Type A. 
$v_0= v_k$, 
$d_{G}(v_0) \geq 3$ and 
$d_{G}(v_i) = 2$ for 
$0 \lt i \lt k$.
Type B. 
$v_0 \neq v_k$, 
$d_{G}(v_0)\geq 3$, 
$d_{G}(v_k) \geq 3$ and 
$d_{G}(v_i) = 2$ for 
$0 \lt i \lt k$.
Two types of internal path.
In Spectral Graph Theory, it is well-known that subdividing an edge can change the spectral radius of the adjacency matrix (see Lemma 2.4). We here provide a novel conclusion for the Laplacian matching indices of graphs (see Lemma 2.5).
Lemma 2.4. ([Reference Hoffman, Smith and Fiedler10])
Let 
$uv$ be an edge of the connected graph 
$G$ and let 
$G_{uv}$ be obtained from 
$G$ by subdividing 
$uv$.
(i) If
$uv$ is not in an internal path and
$G\neq C_{n}$, then
$\rho_{A}(G_{uv}) \gt \rho_{A}(G).$(ii) If
$uv$ is in an internal path and
$G\neq F_{s,1}$, then
$\rho_{A}(G_{uv}) \lt \rho_{A}(G).$
Lemma 2.5. Let 
$uv$ be an edge of the connected graph 
$G$ and let 
$G_{uv}$ be obtained from 
$G$ by subdividing 
$uv$.
(i) If
$uv$ is not in an internal path or
$uv$ is in an internal path of Type A, then
\begin{equation*}\lambda(G_{uv}) \gt \lambda(G).\end{equation*}(ii) If
$uv$ is a cut edge in an internal path of Type B, then
\begin{equation*}\lambda(G_{uv}) \lt \lambda(G).\end{equation*}
Proof. (i) If 
$uv$ is not in an internal path and if 
$G\neq C_{n}$, then 
$G$ is a proper subgraph of 
$G_{uv}$, and so 
$\lambda(G_{uv}) \gt \lambda(G)$ by Lemma (ii). If 
$G=C_{n}$, then 
$S(G)=C_{2n}$ and 
$S(G_{uv})=C_{2n+2}$. Note that the path-tree 
$T(C_{2n},u)=P_{4n-1}$ is a proper subgraph of path-tree 
$T(C_{2n+2},u)=P_{4n+3}$. By Lemma (i) we get 
$\rho_A(T(C_{2n+2},u)) \gt \rho_A(T(C_{2n},u))$, and thus 
$\lambda(G_{uv}) \gt \lambda(G)$ by Corollary 2.2.
If 
$uv$ is in an internal path of Type A, as labelled in Fig. 1, we get 
$T(S(G),v_{0}) \subset T(S(G_{uv}),v_{0})$. Hence, 
$\lambda(G_{uv}) \gt \lambda(G)$ follows by Lemma (i) and Corollary 2.2 again.
(ii) Since 
$uv$ is a cut edge in an internal path of Type B, all edges in this internal path are cut edges. Hence, the path-tree 
$T(S(G_{uv}),v_{0})$ is obtained from 
$T(S(G),v_{0})$ by subdividing some edges of the internal path, as labelled in Type B of Fig. 1. Then 
$\lambda(G_{uv}) \lt \lambda(G)$ by Lemma 2.4(ii) and Corollary 2.2.
We next derive a formula for computing Laplacian matching polynomials. For a subgraph 
$H \subseteq G$, the Laplacian sub-matching polynomial of 
$G$ w.r.t. 
$H$ is defined as [Reference Wan, Wang and Mohammadian17]:
\begin{equation*}\mathscr{L}_{G}(H,x)=\sum_{M\in \mathcal{M}(H)}(-1)^{|M|}\prod_{v\in V(H)\backslash V(M)}\left(x-d_{G}(v)\right).\end{equation*} In particular, 
$\mathscr{L}_{G}(G,x)=\mathscr{L}(G,x)$. It is easy to check the fact that for two vertex-disjoint graphs 
$G_{1}, G_{2}\subseteq G$, we have
\begin{equation*}L_{G}(G_1 \cup G_2,x)=L_{G}(G_1,x)L_{G}(G_2,x).\end{equation*} We will label 
$\mathscr{L}_{G}(H,x)$ as 
$\mathscr{L}_{G}(H)$ if it is clear from the text. The following lemma will be used in the sequel.
Lemma 2.6. Let 
$H \subseteq G$ and 
$v\in V(H)$. Then
\begin{equation}
\mathscr{L}_{G}(H,x)=\left(x-d_{G}(v)\right)\mathscr{L}_{G}(H-v,x)-\sum_{u\in N_{H}(v)}\mathscr{L}_{G}(H-v-u,x),
\end{equation} where 
$N_{H}(v)$ is the set of all vertices in 
$V(H)$ adjacent to 
$v$.
Proof. We partition 
$M\in \mathcal{M}(H)$ into two parts depending on whether 
$v\in V(M)$ or not. Then
\begin{align*}
\mathscr{L}_{G}(H,x)=&\sum_{M\in \mathcal{M}(H)}(-1)^{|M|}\prod_{w\in V(H)\backslash V(M)}\left(x-d_{G}(w)\right)\\[2mm]
=&\sum_{\substack{M\in \mathcal{M}(H)\\v\notin V(M)}}(-1)^{|M|}\prod_{w\in V(H)\backslash V(M)}\left(x-d_{G}(w)\right)\\
&+\sum_{\substack{M\in \mathcal{M}(H)\\v\in V(M)}}(-1)^{|M|}\prod_{w\in V(H)\backslash V(M)}\left(x-d_{G}(w)\right)\\[2mm]
=&\left(x-d_{G}(v)\right)\sum_{M\in \mathcal{M}(H-v)}(-1)^{|M|}\prod_{w\in V(H-v)\backslash V(M)}\left(x-d_{G}(w)\right)\\[2mm]
&+\sum_{u\in N_{H}(v)}{(-1)\sum_{M\in \mathcal{M}(H-v-u)}(-1)^{|M|}\prod_{w\in V(H-v-u)\backslash V(M)}\left(x-d_{G}(w)\right)}\\[2mm]
=&\left(x-d_{G}(v)\right)\mathscr{L}_{G}(H-v,x)-\sum_{u\in N_{H}(v)}\mathscr{L}_{G}(H-v-u,x).
\end{align*}This completes the proof of this lemma.
Let 
$P_{n}$ be a path of order 
$n$ with its vertices labelled as 
$v_{1}v_{2}\cdots v_{n-1}v_{n}$. Set 
$\psi_{n}(x) = \mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1},x)$ and 
$\phi_{n}(x) = \mathscr{L}_{P_{n+2}}(P_{n+2}-v_{1}-v_{n+2},x)$.
Lemma 2.7. Under the introduced notation, the following results hold:
(i)
$\mathscr{L}(P_{n+1},x)=(x-2)\mathscr{L}(P_{n},x)-\mathscr{L}(P_{n-1},x).$(ii)
$x\psi_{n}(x)=\mathscr{L}(P_{n+1},x)+\mathscr{L}(P_{n},x).$(iii)
$x\phi_{n}(x)=\mathscr{L}(P_{n+1},x).$
Proof. Due to (2), we get
\begin{align}
\qquad\quad\mathscr{L}(P_{n+1},x)&=(x-1)\mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1},x)-\mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1}-v_{2},x)\nonumber\\
&=(x-2)\mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1},x)-\mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1}-v_{2},x)\nonumber\\
&\quad +\mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1},x)\nonumber\\
&=\mathscr{L}_{P_{n+2}}(P_{n+2}-v_{1},x)+\mathscr{L}_{P_{n+1}}(P_{n+1}-v_{1},x)\nonumber\\
&=\psi_{n+1}(x)+\psi_{n}(x),
\end{align}
\begin{align}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\psi_{n+1}(x)&=\mathscr{L}_{P_{n+2}}(P_{n+2}-v_{1},x)\qquad\qquad\qquad\nonumber\\
&=(x-1)\mathscr{L}_{P_{n+2}}(P_{n+2}-v_{1}-v_{n+2},x) \nonumber\\
&\quad -\mathscr{L}_{P_{n+2}}(P_{n+2}-v_{1}-v_{n+1}-v_{n+2},x)\nonumber\\
&=(x-1)\phi_{n}(x)-\phi_{n-1}(x)\nonumber\\
&=(x-2)\phi_{n}(x)-\phi_{n-1}(x)+\phi_{n}(x)\nonumber\\
&=\phi_{n+1}(x)+\phi_{n}(x).
\end{align}In the light of (2), (3) and (4),
\begin{equation}
x\psi_{n}(x)=x\psi_{n}(x)+\mathscr{L}(P_{n},x)-\left(\psi_{n}(x)+\psi_{n-1}(x)\right)
=\mathscr{L}(P_{n+1},x)+\mathscr{L}(P_{n},x),
\end{equation}and
\begin{align*}
\mathscr{L}(P_{n+1},x)&=(x-1)\psi_{n}(x)-\psi_{n-1}(x) \nonumber\\
&=(x-1)\left((x-2)\psi_{n-1}(x)-\psi_{n-2}(x)\right)-\psi_{n-1}(x)\nonumber\\
&=(x-2)\mathscr{L}(P_{n},x)-\left(\psi_{n-1}(x)+\psi_{n-2}(x)\right)\nonumber\\
&=(x-2)\mathscr{L}(P_{n},x)-\mathscr{L}(P_{n-1},x).\nonumber
\end{align*}
\begin{align*}
x\phi_{n+1}(x)-\mathscr{L}(P_{n+2},x)=\mathscr{L}(P_{n+1},x)-x\phi_{n}(x),
\end{align*}which is equivalent to
\begin{equation*}x\phi_{n}(x)-\mathscr{L}(P_{n+1},x)=(-1)^{n-1}\left(x\phi_{1}(x)-\mathscr{L}(P_{2},x)\right)=0, \;\;{\mbox {i.e.,}}\;\; x\phi_{n}(x) = \mathscr{L}(P_{n+1},x).\end{equation*}This finishes our proof.
Lemma 2.8. Let 
$a=\frac{x-2+\sqrt{x^{2}-4x}}{2}$ and 
$b=\frac{x-2-\sqrt{x^{2}-4x}}{2}$. Then
(i)
$\mathscr{L}(P_{n},x)=\frac{x}{\sqrt{x^{2}-4x}}(a^{n}-b^{n}).$(ii)
$\psi_{n}(x)=\frac{1}{\sqrt{x^{2}-4x}}(a^{n+1}+a^{n}-b^{n+1}-b^{n}).$(iii)
$\phi_{n}(x)=\frac{1}{\sqrt{x^{2}-4x}}(a^{n+1}-b^{n+1}).$
Proof. For (i), we employ induction on 
$n$. It is easy to verify the result for 
$n=1,2$. Suppose that the result holds if 
$1\leq k\leq n$ and 
$n\geq3$. When 
$k=n+1$, in terms of Lemma 2.7(i) we get
\begin{align*}
\mathscr{L}(P_{n+1},x)&=(x-2)\mathscr{L}(P_{n},x)-\mathscr{L}(P_{n-1},x)\\
&=\frac{x}{\sqrt{x^{2}-4x}}\left((x-2)(a^{n}-b^{n})-(a^{n-1}-b^{n-1})\right)\\
&=\frac{x}{\sqrt{x^{2}-4x}}\left((x-2)a^{n}-a^{n-1}-\left((x-2)b^{n}-b^{n-1}\right)\right)\\
&=\frac{x}{\sqrt{x^{2}-4x}}\left(a^{n+1}-b^{n+1}\right).
\end{align*}Further, (ii) and (iii) follow from Lemma 2.7(ii) and (iii).
Remark 2.9. By Mohammadian’s result in (1), Lemma 2.8(i) can be directly deduced by [[Reference Guo7], Lemma 3.3]. The same Mohammadian’s result, together with 
$\lim_{i \to \infty}\rho_{L}(P_{i})$ noted in [Reference Guo7], gives an alternative proof to the subsequent Corollary 2.10(i).
Corollary 2.10. Let 
$\{G_{i}\}_{i=1}^{\infty}$ be a graph sequence.
(i) If
$G_{i}=P_{i}$, then
$\lim\limits_{i \to \infty}\lambda(P_{i})=4$.(ii) If
$G_{i}=C_{i}$, then
$\lim\limits_{i \to \infty}\lambda(C_{i})=4$.
Proof. From 
$P_{i}\subseteq P_{i+1}$ and 
$\Delta(P_{i})\leq2$ it follows by Lemma (ii) that 
$\lambda(P_{i}) \lt \lambda(P_{i+1}) \lt 4$. Since 
$P_{i}$ is a tree, by (1) we obtain 
$\lambda(P_{i})=\rho_L(P_{i})=4\cos^{2}\frac{\pi}{2i}$ (see [Reference Anderson, William and Morley1]) showing 
$\lim\limits_{i \to \infty}\lambda(P_{i})=4$. In terms of Lemma (ii) again, 
$\lambda(C_{i})=2+2\cos \frac{\pi}{2i} \lt 4$ and thus 
$\lim\limits_{i \to \infty}\lambda(C_{i})=4$.
We now pay attention to the limit points of the Laplacian matching indices of graphs.
Lemma 2.11. Let 
$G$ be a connected graph and 
$v\in V(G)$. 
$Q_{v}(G;kP_{n})$ is obtained from vertex-disjoint graphs 
$G$ and 
$k$ paths 
$P^{}_{n}$ by adding 
$k$ new edges 
$vv_{i}$ with 
$i=1,\cdots, k$, respectively, where 
$v_{i}$ is an end vertex of 
$i$-th path 
$P_{n}$. Then
(i)
$\lim\limits_{n \to \infty}\lambda(Q_{v}(G;kP_{n}))=r$ exists and
$r\geq 4$.(ii) If
$r \gt 4$, then
$r$ is the largest positive root of the equation:
\begin{equation*}
\frac{x+\sqrt{x^{2}-4x}}{2kx}\mathscr{L}(G,x)-\mathscr{L}_{G}(G-v,x)=0.
\end{equation*}
Proof. Since 
$Q_{v}(G;kP_{n})\subset Q_{v}(G;kP_{n+1})$, then 
$\lim\limits_{n \to \infty}\lambda(Q_{v}(G;kP_{n}))=r$ exists by Lemma (ii), and thus 
$r\geq 4$ by 
$P_{n}\subset Q_{v}(G;kP_{n})$ and Corollary 2.10(i).
For simplicity, set 
$G'=Q_{v}(G;kP_{n})$. In the light of the equality in (2),
\begin{align*}
\mathscr{L}(G')&=(x-d_{G}(v)-k)\mathscr{L}_{G'}(G'-v)-\sum_{u\in N_{G'}(v)}\mathscr{L}_{G'}(G'-v-u)\\
&=(x-d_{G}(v))\mathscr{L}_{G}(G-v)\psi^{k}_{n}(x)-\sum_{u\in N_{G}(v)}\mathscr{L}_{G}(G-v-u)\psi^{k}_{n}(x)\\
&\quad-k\mathscr{L}_{G}(G-v)\psi^{k}_{n}(x)-k\mathscr{L}_{G}(G-v)\psi_{n-1}(x)\psi^{k-1}_{n}(x)\\[2mm]
&=\mathscr{L}(G)\psi^{k}_{n}(x)-k\mathscr{L}_{G}(G-v)\psi^{k}_{n}(x)-k\mathscr{L}_{G}(G-v)\psi_{n-1}(x)\psi^{k-1}_{n}(x)\\[2mm]
&= \mathscr{L}^{k}(P_{n})\left(\frac{\psi^{k}_{n}(x)}{\mathscr{L}^{k}(P_{n})}\left[\mathscr{L}(G)-k\mathscr{L}_{G}(G-v)\right]-k\frac{\psi_{n-1}(x)\psi^{k-1}_{n}(x)}{\mathscr{L}^{k}(P_{n})}\mathscr{L}_{G}(G-v)\right).
\end{align*} For 
$x \gt 4$, easily to verify that 
$\frac{b}{a} \lt 1$. By Lemma 2.8 we have
\begin{align*}
&\lim_{n \to \infty}\left(\frac{\psi^{k}_{n}(x)}{\mathscr{L}^{k}(P_{n})}\left[\mathscr{L}(G)-k\mathscr{L}_{G}(G-v)\right]-k\frac{\psi_{n-1}(x)\psi^{k-1}_{n}(x)}{\mathscr{L}^{k}(P_{n})}\mathscr{L}_{G}(G-v)\right)\\[2mm]
&\quad =\left(\frac{a+1}{x}\right)^{k}\left(\mathscr{L}(G)-k\mathscr{L}_{G}(G-v)\right)-k\frac{b+1}{x}\left(\frac{a+1}{x}\right)^{k}\mathscr{L}_{G}(G-v)\\[2mm]
&\quad=k\left(\frac{a+1}{x}\right)^{k-1}\left[\frac{a+1}{x}\left(\dfrac{1}{k}\mathscr{L}(G)-\mathscr{L}_{G}(G-v)\right)-\frac{b+1}{x}\mathscr{L}_{G}(G-v)\right]\\[2mm]
&\quad=k\left(\frac{a+1}{x}\right)^{k-1}\left(\frac{x+\sqrt{x^{2}-4x}}{2kx}\mathscr{L}(G)-\mathscr{L}_{G}(G-v)\right).
\end{align*} Since 
$r \gt 4$, we conclude that 
$r$ is the largest positive root of the equation:
\begin{equation*}
\frac{x+\sqrt{x^{2}-4x}}{2kx}\mathscr{L}(G,x)-\mathscr{L}_{G}(G-v,x)=0.
\end{equation*}This finishes the proof.
Next, we apply the above lemma to a class of specific graphs. Let 
$T_{i,j,k}$ be the tree with exactly one vertex 
$u$ of maximum degree three such that 
$T_{i,j,k} - u = P_i \cup P_j \cup P_k$, which is shown in Fig. 2.
The tree 
$T_{i,j,k}$ 
$(i,j,k\geq1)$.
Corollary 2.12. For the tree 
$T_{1,i,i}$, 
$\lim\limits_{i \to \infty}\lambda(T_{1,i,i})=\tau$.
Proof. Note that 
$T_{1,i,i} \subset T_{1,i+1,i+1}$, then 
$\lambda(T_{1,i,i}) \lt \lambda(T_{1,i+1,i+1})$ and the sequence 
$\{\lambda(T_{1,i,i})\}_{i=1}^{\infty}$ is strictly increasing from Lemma (ii). Hence, 
$\lim\limits_{i \to \infty}\lambda(T_{1,i,i})=r \gt 4$ exists.
Let 
$v$ be an end vertex of 
$P_{2}$. In the light of Lemma 2.11(ii), then 
$r$ is the largest root of the equation:
\begin{equation*}\frac{x+\sqrt{x^{2}-4x}}{4x}\mathscr{L}(P_{2},x)-\mathscr{L}_{P_{2}}(P_{2}-v,x)=0.\end{equation*} Since 
$\mathscr{L}(P_{2},x)=x^{2}-2x$ and 
$\mathscr{L}_{P_{2}}(P_{2}-v,x)=x-1$, the equation is equivalent to
\begin{equation*}\frac{x+\sqrt{x^{2}-4x}}{4x}(x^{2}-2x)-(x-1)=0, \; {\mbox{i.e.},}\;x^{3}-6x^{2}+8x-4=0,\end{equation*} whose largest root is 
$\tau$. Hence, the result follows.
Proposition 2.13. Let 
$G_{1}$ and 
$G_{2}$ be two vertex-disjoint connected graphs with 
$u \in V(G_1)$ and 
$v \in V(G_2)$. 
$\tilde{Q}_{uv}(G_{1},G_{2};P_{n})$ is the graph obtained from 
$G_{1}$, 
$G_{2}$ and a new path 
$P_{n}$: 
$v_{1}v_{2}\cdots v_{n-1}v_{n}$ by adding two new edges 
$uv_{1}$ and 
$vv_{n}$, respectively. Then
\begin{equation*}
\lim_{n \to \infty}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))=\max \left\{\lim_{n \to \infty}\lambda(Q_{u}(G_{1};P_{n})),\lim_{n \to \infty}\lambda(Q_{v}(G_{2};P_{n}))\right\}.
\end{equation*}Proof. From Lemma 2.11(i), both 
$\lim\limits_{n \to \infty}\lambda(Q_{u}(G_{1};P_{n}))$ and 
$\lim\limits_{n \to \infty}\lambda(Q_{v}(G_{2};P_{n}))$ exist.
If 
$uv_{1}v_{2}\cdots v_{n}v$ is not in an internal path, then either 
$Q_{u}(G_{1};P_{n})$ or 
$Q_{v}(G_{2};P_{n})$ is a path. Obviously, 
$\tilde{Q}_{uv}(G_{1},G_{2};P_{n})\subset \tilde{Q}_{uv}(G_{1},G_{2};P_{n+1})$, and thus we obtain that 
$\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n})) \lt \lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n+1}))$ by Lemma (ii). Therefore, the sequence 
$\{\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))\}_{n=1}^{\infty}$ is bounded, and so
\begin{equation*}
\lim_{n \to \infty}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))=\max \left\{\lim_{n \to \infty}\lambda(Q_{u}(G_{1};P_{n})),\lim_{n \to \infty}\lambda(Q_{v}(G_{2};P_{n}))\right\}.
\end{equation*} We next consider that 
$uv_{1}v_{2}\cdots v_{n}v$ is in an internal path. Since 
$uv_{1}$ is a cut edge, it follows from Lemma 2.5(ii) that
\begin{equation*}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n})) \gt \lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n+1})).\end{equation*} Hence, 
$\lim_{n \to \infty}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))$ exists. Note that 
$\Delta(Q_{u}(G_{1};P_{n})),\Delta(Q_{v}(G_{2};P_{n}))\geq 3$ and 
$Q_{u}(G_{1};P_{n}),Q_{v}(G_{2};P_{n})\subset \tilde{Q}_{uv}(G_{1},G_{2};P_{n})$. Along with Lemmas (ii) and 2.11, we have
\begin{equation*}
\lim_{n \to \infty}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))\geq \lim_{n \to \infty}\lambda(Q_{u}(G_{1};P_{n})) \gt 4
\end{equation*}and
\begin{equation*}
\lim_{n \to \infty}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))\geq \lim_{n \to \infty}\lambda(Q_{v}(G_{2};P_{n})) \gt 4.
\end{equation*} Set 
$G = \tilde{Q}_{uv}(G_{1},G_{2};P_{n}), G_{1}=Q_{u}(G_{1};P_{n})$ and 
$G_{2}=Q_{v}(G_{2};P_{n})$. In the light of (2),
\begin{align*}
\mathscr{L}(G)&=(x-d_{G_{1}}(u)-1)\mathscr{L}_{G}(G-u)-\sum_{w\in N_{G}(u)}\mathscr{L}_{G}(G-u-w)\\
&=(x-d_{G_{1}}(u)-1)\mathscr{L}_{G_{1}}(G_{1}-u)\mathscr{L}_{G}(Q_{v}(G_{2};P_{n}))\\
&\quad -\sum_{w\in N_{G_{1}}(u)}\mathscr{L}_{G_{1}}(G_{1}-u-w)\mathscr{L}_{G}(Q_{v}(G_{2};P_{n}))\\
&\quad-\mathscr{L}_{G_{1}}(G_{1}-u)\mathscr{L}_{G}(Q_{v}(G_{2};P_{n})-v_{1})
\end{align*}
\begin{align*}
&=\left(\mathscr{L}(G_{1})-\mathscr{L}_{G_{1}}(G_{1}-u)\right)\mathscr{L}_{G}(Q_{v}(G_{2};P_{n}))\\
&\quad -\mathscr{L}_{G_{1}}(G_{1}-u)\mathscr{L}_{G}(Q_{v}(G_{2};P_{n})-v_{1})\\[2mm]
&=\left(\mathscr{L}(G_{1})-\mathscr{L}_{G_{1}}(G_{1}-u)\right)[\phi_{n}(x)\left(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right)\\
&\quad -\phi_{n-1}(x)\mathscr{L}_{G_{2}}(G_{2}-v)]\\[2mm] &\quad-\mathscr{L}_{G_{1}}(G_{1}-u)\left[\phi_{n-1}(x)\left(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right)-\phi_{n-2}(x)\mathscr{L}_{G_{2}}(G_{2}-v)\right]\\[2mm]
&=\left[\left(\mathscr{L}(G_{1})-\mathscr{L}_{G_{1}}(G_{1}-u)\right)\left(\frac{\phi_{n}(x)}{\mathscr{L}(P_{n})}\left(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right)\right.\right.\\
&\quad \left. -\frac{\phi_{n-1}(x)}{\mathscr{L}(P_{n})}\mathscr{L}_{G_{2}}(G_{2}-v)\right)-\mathscr{L}_{G_{1}}(G_{1}-u)\\[2mm]
&\quad\times \left.\left(\frac{\phi_{n-1}(x)}{\mathscr{L}(P_{n})}\left(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right) -\frac{\phi_{n-2}(x)}{\mathscr{L}(P_{n})}\mathscr{L}_{G_{2}}(G_{2}-v)\right)\right]\mathscr{L}(P_{n})\\[2mm]
&\triangleq \mathscr{L}(P_{n})f_{n}(x).
\end{align*} For 
$x \gt 4$, from Lemma 2.8 we have
\begin{align*}
\lim_{n \to \infty}f_{n}(x)&=\left(\mathscr{L}(G_{1})-\mathscr{L}_{G_{1}}(G_{1}-u)\right)\left(\lim_{n \to \infty}\frac{\phi_{n}(x)}{\mathscr{L}(P_{n})}\left(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right)\right.\\
&\quad-\left.\lim_{n \to \infty}\frac{\phi_{n-1}(x)}{\mathscr{L}(P_{n})}\mathscr{L}_{G_{2}}(G_{2}-v)\right)\\
&\quad-\mathscr{L}_{G_{1}}(G_{1}-u)\left(\lim_{n \to \infty}\frac{\phi_{n-1}(x)}{\mathscr{L}(P_{n})}\left(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right) \right.\\
&\quad\left. -\lim_{n \to \infty}\frac{\phi_{n-2}(x)}{\mathscr{L}(P_{n})}\mathscr{L}_{G_{2}}(G_{2}-v)\right)\\[2mm]
&=(\mathscr{L}(G_{1})-\mathscr{L}_{G_{1}}(G_{1}-u))\left(\frac{a}{x}(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v))
-\frac{1}{x}\mathscr{L}_{G_{2}}(G_{2}-v)\right)\\[2mm]
&\quad- \mathscr{L}_{G_{1}}(G_{1}-u)\left(\frac{1}{x}(\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v))
-\frac{b}{x}\mathscr{L}_{G_{2}}(G_{2}-v)\right)\\[2mm]
&=\left(\frac{x+\sqrt{x^{2}-4x}}{2x}\mathscr{L}(G_{1})-\mathscr{L}_{G_{1}}(G_{1}-u)\right)\\
&\quad \times \left(\frac{x+\sqrt{x^{2}-4x}}{2x}\mathscr{L}(G_{2})-\mathscr{L}_{G_{2}}(G_{2}-v)\right)\\[2mm]
&\triangleq f(x).
\end{align*} Since 
$\lim\limits_{n \to \infty}\lambda(G) \gt 4$, we conclude that 
$\lim\limits_{n \to \infty}\lambda(G)$ is the largest positive root of the equation 
$f(x)=0$. Thereby, it follows from Lemma 2.11 that
\begin{equation*}
\lim_{n \to \infty}\lambda(\tilde{Q}_{uv}(G_{1},G_{2};P_{n}))=\max \left\{\lim_{n \to \infty}\lambda(Q_{u}(G_{1};P_{n})),\lim_{n \to \infty}\lambda(Q_{v}(G_{2};P_{n}))\right\}.
\end{equation*}This finishes our proof.
Let 
$H_{g,t}$ be the lollipop graph shown in Fig. 3. At the end of this section, we focus on the limit points of the largest Laplacian matching root of 
$H_{g,t}$ when 
$g$ and 
$t$ tend to infinity, respectively.
The graph 
$H_{g,t}$ 
$(g\geq3, t\geq1)$.
Proposition 2.14. Let 
$H_{g,t}$ be defined as above. Then
\begin{equation*}\lim_{g \to \infty}\lambda(H_{g,1})=\tau.\end{equation*}Proof. Due to Lemma 2.5(i), 
$\{\lambda(H_{g,1})\}_{g=3}^{\infty}$ is a strictly increasing sequence. Along with Lemma (ii) we get that 
$\lim\limits_{g \to \infty}\lambda(H_{g,1})=r \gt 4$ exists.
For simplicity, let 
$H'=H_{g,1}$ and 
$v$ be the vertex of 
$C_{g}$ with 
$d_{H'}(v)=3$. In the light of (2),
\begin{align*}
\mathscr{L}(H')&=(x-3)\mathscr{L}_{H'}(H'-v)-\sum_{u\in N_{H'}(v)}\mathscr{L}_{H'}(H'-v-u)\\
&=(x-3)\phi_{g-1}(x)\psi_{1}(x)-2\phi_{g-2}(x)\psi_{1}(x)-\phi_{g-1}(x)\\[2mm]
&=(x^2-4x+2)\phi_{g-1}(x)-2(x-1)\phi_{g-2}(x)\\[2mm]
&= \mathscr{L}(P_{g})\left[\left(\frac{\phi_{g-1}(x)}{\mathscr{L}(P_{g})}\right)(x^2-4x+2)-2\frac{\phi_{g-2}(x)}{\mathscr{L}(P_{g})}(x-1)\right].
\end{align*} From Lemma 2.8, we have for 
$x \gt 4$
\begin{align*}
&\lim_{g \to \infty}\left[\left(\frac{\phi_{g-1}(x)}{\mathscr{L}(P_{g})}\right)(x^2-4x+2)-2\frac{\phi_{g-2}(x)}{\mathscr{L}(P_{g})}(x-1)\right]\\
=&(x^2-4x+2)\frac{1}{x}-(x-1)\frac{2b}{x}\\
=&\frac{1}{x}\left((x-1)\sqrt{x^2-4x}-x\right),
\end{align*} which implies that 
$\lim\limits_{g \to \infty}\lambda(H_{g,1})=r$ is the largest positive root of the equation:
\begin{equation*}(x-1)\sqrt{x^2-4x}-x=0, \; {\mbox{i.e.},}\;x^{3}-6x^{2}+8x-4=0.\end{equation*}Thus, we finally conclude that
\begin{equation*}\lim_{g \to \infty}\lambda(H_{g,1})=\tau.\end{equation*}This finishes the proof.
Proposition 2.15. Let 
$H_{g,t}$ be defined as above. Then for every 
$g\geq 3$,
\begin{equation*}\lim_{t \to \infty}\lambda(H_{g,t}) \gt \tau.\end{equation*}Proof. Since 
$H_{g,t} \subset H_{g,t+1}$, by Lemma (ii) we get 
$\lambda(H_{g,t}) \lt \lambda(H_{g,t+1})$, and thus the sequence 
$\{\lambda(H_{g,t})\}_{t=1}^{\infty}$ is strictly increasing. Hence, 
$\lim\limits_{t \to \infty}\lambda(H_{g,t})=r_{g} \gt 4$ exists by Lemma (ii) again. On the other hand, from Lemma 2.5(i) it follows that 
$\lambda(H_{g+1,t}) \gt \lambda(H_{g,t})$ which indicates
\begin{equation}
\lim_{t \to \infty}\lambda(H_{g,t}) \geq \lim_{t \to \infty}\lambda(H_{3,t}).
\end{equation} We now consider the sequence 
$\{\lambda(H_{3,t})\}_{t=1}^{\infty}$. Let 
$v\in V(C_{3})$ with 
$d_{H_{3,t}}(v)=3$. In the light of Lemma 2.11 (ii), 
$r_{3}$ is the largest root of the equation:
\begin{equation}
\frac{x+\sqrt{x^{2}-4x}}{2x}\mathscr{L}(C_{3},x)-\mathscr{L}_{C_{3}}(C_{3}-v,x)=0.
\end{equation} Substituting 
$\mathscr{L}(C_{3},x) = x^{3}-6x^{2}+9x-2$ and 
$\mathscr{L}_{C_{3}}(C_{3}-v,x) = x^{2}-4x+3$ into (7), then
\begin{equation*}\frac{x+\sqrt{x^{2}-4x}}{2x}(x^{3}-6x^{2}+9x-2)-(x^{2}-4x+3)=0,\end{equation*}which is equivalent to
\begin{equation*}x^{5}-10x^{4}+34x^{3}-46x^{2}+21x-4=0.\end{equation*} Thereby, 
$\lim\limits_{t \to \infty}\lambda(H_{3,t})=r_{3}\approx 4.47 \gt \tau$, and hence the result follows from (6).
3. Main results
We are now in the stage to obtain the main results, where the method used in Theorem 3.1 is inspired by Hoffman [Reference Hoffman9] and Guo [Reference Guo7].
Theorem 3.1 Let 
$\beta_{0}=1$ and 
$\beta_{n}$ be the largest positive root of
\begin{equation*}h_{n}(x)=x^{n+1}-(1+x+...+x^{n-1})(\sqrt{x}+1)^{2}.\end{equation*} And let 
$\alpha_{n}=2+\beta_{n}^{\frac{1}{2}}+\beta_{n}^{-\frac{1}{2}}$. Then
\begin{equation*}4=\alpha_{0} \lt \alpha_{1} \lt \alpha_{2} \lt \cdots\end{equation*} are all the limit points of the Laplacian matching indices of graphs smaller than 
$\lim\limits_{n \to \infty}\alpha_{n}=\tau$.
Proof. Obviously, we suppose each one in the graph sequence 
$\{G_{i}\}_{i=1}^{\infty}$ to be connected from the fact that 
$\lambda(G) = \max\{\lambda(\tilde{G}_1),\lambda(\tilde{G}_2)\}$ for 
$G = \tilde{G}_1 \cup \tilde{G}_2$. Then for 
$i\neq j$, set 
$\lambda(G_{i})\neq\lambda(G_{j})$ and 
$\lim\limits_{i \to \infty}\lambda(G_i)=r \lt \tau$. If 
$\Delta(G_{i}) \geq 4$, by Lemma (ii), we get 
$\lambda(G_{i})\geq 5$ and so 
$r\geq 5$, a contradiction. Thus, 
$\Delta(G_{i})\leq 3$. If 
$\Delta(G_{i})=2$ for infinitely many 
$i's$, then 
$G_{i}$ is a path or a cycle. It follows from Corollary 2.10 that
\begin{equation*}\lim_{i \to \infty}\lambda(G_i)=\lim_{i \to \infty}\lambda(P_{i})=\lim_{i \to \infty}\lambda(C_{i})=4.\end{equation*} Thereby, in what follows, we always set 
$\Delta(G_{i})=3$.
Let 
$\mathcal{S}=\{g_{i}\,\vert\, i\geq1\}$ where 
$g_{i}$ is the girth of 
$G_{i}$. The following cases are to be considered:
Case 1. 
$\mathcal{S}$ is unbounded. Assume that the cycle 
$C_{g_i} \subseteq G_i$ with 
$g_i \gt g_j$ for 
$i \gt j$. Since 
$\Delta(G_i)=3$, then 
$H_{g_i,1} \subseteq G_i$ and so 
$\lambda(G_i)\geq \lambda (H_{g_i,1})$. Along with Proposition 2.14 we get
\begin{equation*}\lim_{i \to \infty}\lambda(G_i)\geq \lim_{i \to \infty}\lambda(H_{g_i,1})=\tau,\end{equation*}a contradiction.
Case 2. 
$\mathcal{S}$ is bounded and there are infinitely many 
$G_{i}$’s which are not trees. If 
$C_{g} \subset G_{i}$, by 
$\Delta(G_{i})=3$ we obtain 
$H_{g,k_{i}} \subseteq G_{i}$ with 
$k_{i} \gt k_{j}$ for 
$i \gt j$. Similarly, by Proposition 2.15 it follows that
\begin{equation*}\lim_{i \to \infty}\lambda(G_{i})\geq \lim_{i \to \infty}\lambda(H_{g,k_{i}}) \gt \tau,\end{equation*}a contradiction again.
Case 3. Infinitely many 
$G_{i}$’s are trees. If 
$G_{i}$ has at least three vertices of degree three, then 
$R_{p_{i},q_{i}}\subseteq G_{i}$ (see Fig. 4). When 
$k_{i}$ is sufficiently large, by Lemma 2.5(ii) we get 
$\lambda(G_{i})\geq \lambda(R_{p_{i},q_{i}})\geq \lambda(T_{1,k_{i},k_{i}})$. Due to Corollary 2.12, we obtain that
\begin{equation*}
\lim_{i \to \infty}\lambda(G_{i})\geq \lim_{i \to \infty}\lambda(T_{1,k_{i},k_{i}})=\tau,
\end{equation*}a contradiction.
The graph 
$R_{p,q}$ 
$(p,q\geq1)$.
If 
$G_{i}$ has two vertices of degree three and their distance is bounded, then 
$F_{s,r_{i}}\subseteq G_{i}$ (see Fig. 5) with 
$r_{i} \gt r_{j}$ for 
$i \gt j$. In terms of Lemma 2.5(ii), for a sufficiently large integer 
$k_{i}$, it follows that 
$\lambda(G_{i})\geq \lambda(F_{s,r_{i}})\geq \lambda(T_{1,k_{i},k_{i}})$. Along with Corollary 2.12, we have
\begin{equation*}
\lim_{i \to \infty}\lambda(G_{i})\geq \lim_{i \to \infty}\lambda(T_{1,k_{i},k_{i}})=\tau,
\end{equation*}a contradiction again.
The graph 
$F_{s,r_i}$ 
$(s \gt 1,r_i\geq1)$.
If 
$G_{i}$ has two vertices of degree three and the distance between them is unbounded, then by Proposition 2.13 we can construct a tree sequence 
$\{T_{k,n,m_{i}}\}_{i=1}^{\infty}$ with 
$m_{i} \gt m_{j}$ for 
$i \gt j$ such that 
$\lim_{i \to \infty}\lambda(T_{k,n,m_{i}})=\lim_{i \to \infty}\lambda(G_{i})$. Without loss of generality, set 
$m_i \geq n \geq k \geq 1$. If 
$k \geq 2$, then 
$T_{2,2,2} \subseteq T_{k,n,m_i}$, and thus 
$\lambda(T_{2,2,2})=3+\sqrt{2} \gt \tau$, a contradiction. So, 
$k=1$ and 
$T_{1,n,m_i}$ is shown in Fig. 6.
The graph 
$T_{1,n,m_{i}}$.
As stated above, the limit points we want are the numbers 
$\lim\limits_{i \to \infty}\lambda(T_{1,n,m_{i}})=\alpha_{n}$. Due to Lemmas (ii) and 2.11, 
$\alpha_{n}$ is the largest root of the equation:
\begin{equation*}
\frac{x+\sqrt{x^{2}-4x}}{2x}\mathscr{L}(P_{n+2},x)-\psi_{1}(x)\psi_{n}(x)=0.
\end{equation*} Let 
$x=\gamma +\frac{1}{\gamma}+2$ and 
$\gamma^{2}=z$. From Lemma 2.8 we arrive at
\begin{align*}
&\frac{x+\sqrt{x^{2}-4x}}{2x}\mathscr{L}(P_{n+2},x)-\psi_{1}(x)\psi_{n}(x)\\
=&\frac{x+\sqrt{x^{2}-4x}}{2x}\mathscr{L}(P_{n+2},x)-(x-1)\psi_{n}(x)\\
=&\frac{x+\sqrt{x^{2}-4x}}{2\sqrt{x^{2}-4x}}(a^{n+2}-b^{n+2})-\frac{1}{\sqrt{x^{2}-4x}}(x-1)(a^{n+1}+a^{n}-b^{n+1}-b^{n})\\
=&(\gamma+1)\frac{\gamma}{\gamma^{2}-1}\bigg(\gamma^{n+2}-\frac{1}{\gamma^{n+2}}\bigg)-\frac{\gamma}{\gamma^{2}-1}\bigg(\gamma+\frac{1}{\gamma}+1\bigg)\bigg(\gamma^{n+1}+\gamma^{n}-\frac{1}{\gamma^{n+1}}-\frac{1}{\gamma^{n}}\bigg)\\
=&\frac{1}{\gamma^{n}}\left[\frac{\gamma}{\gamma-1}\bigg(\gamma^{2n+2}-\frac{1}{\gamma^{2}}\bigg)-\frac{\gamma}{\gamma-1}\bigg(\gamma+\dfrac{1}{\gamma}+1\bigg)\bigg(\gamma^{2n}-\dfrac{1}{\gamma}\bigg)\right]\\
=&\frac{1}{\gamma^{n}}\bigg(\gamma^{2n+2}+\dfrac{\gamma^{2n}+\gamma^{2n+1}-\gamma-1}{1-\gamma}\bigg)\\
=&\frac{1}{\gamma^{n}}\left(\gamma^{2n+2}+\frac{\gamma^{2n}-1}{1-\gamma^{2}}(\gamma+1)^{2}\right)\\
=&\frac{1}{\gamma^{n}}\left(\gamma^{2n+2}-(1+\gamma^{2}+\cdots+\gamma^{2n-2})(\gamma+1)^{2}\right)\\
=&\frac{1}{\gamma^{n}}\left(z^{n+1}-(1+z+\cdots+z^{n-1})(\sqrt{z}+1)^{2}\right).
\end{align*} Since 
$h_{n}(x)-xh_{n-1}(x)=-(\sqrt{x}+1)^{2} \lt 0$ for 
$x\geq 0$, then 
$\beta_{n} \gt \beta_{n-1}$, and hence the sequence 
$\{\alpha_{n}\}_{n=1}^{\infty}$ is strictly increasing. From Corollary 2.12 it follows that 
$\lim\limits_{n \to \infty}\alpha_{n}=\tau$.
This completes the proof.
Theorem 3.2 The first two smallest limit points of the Laplacian matching indices of connected graphs are 
$4$ and 
$2+\sqrt{5}$, respectively. Moreover, for a connected graph 
$G$ with order 
$n$,
(i)
$\lambda(G) \lt 4$ if and only if
$G\in\{P_n | n \geq 1\} \cup \{C_n | n \geq 3\}$.(ii)
$\lambda(G)=4$ if and only if
$G=T_{1,1,1}$.(iii)
$4 \lt \lambda(G)\leq 2+\sqrt{5}$ if and only if
$G\in\{T_{1,1,n-3} | n \geq 5\}$.
Proof. From Theorem 3.1, we know that the smallest limit point of the Laplacian matching indices is 
$\alpha_0 =4$ and the second smallest one is 
$\alpha_{1}=2+\beta_{1}^{\frac{1}{2}}+\beta_{1}^{-\frac{1}{2}}$ with 
$\beta_{1}=\frac{3+\sqrt{5}}{2}$ being the largest positive root of the equation 
$x^{2}-(\sqrt{x}+1)^{2}=0.$ Thus, 
$\alpha_{1}=2+\sqrt{5}$.
We now characterise the connected graphs with 
$\lambda(G) \leq 2+\sqrt{5}$. When 
$G=C_{n}$, by Corollary 2.10(ii) we get 
$\lambda(C_{n})=2+2\cos \frac{\pi}{2n} \lt 4$. In what follows, set 
$G\neq C_{n}$. We claim that 
$G$ must be a tree with 
$\Delta(G)\leq 3$. Otherwise, if 
$\Delta(G) \geq 4$ then 
$\lambda(G) \geq 5$ by Lemma (ii), a contradiction. If 
$G$ is not a tree, then 
$H_{g,1}\subseteq G$ and hence
\begin{equation*}\lambda(G)\geq \lambda(H_{g,1})\geq \lambda(H_{3,1})\approx 4.34 \gt 2+\sqrt{5}\end{equation*}by Lemma 2.5(i), a contradiction.
For 
$\Delta(G)\leq 2$, then 
$G=P_{n}$ with 
$\lambda(P_{n})=4\cos^{2}\frac{\pi}{2n} \lt 4$ by Corollary 2.10(i). For 
$\Delta(G)= 3$, we claim that 
$G$ has exactly one vertex of degree three. Otherwise, 
$G$ has at least two vertices of degree three. So 
$F_{s,1}\subseteq G$. In terms of Lemma 2.5(ii) and Proposition 2.13, the graph sequence 
$\{F_{s,1}\}_{s=1}^{\infty}$ is strictly decreasing and 
$\lim_{s \to \infty}\lambda(F_{s,1})=\lim_{s \to \infty}\lambda(T_{1,1,s})=2+\sqrt{5}.$ Thus, it follows from Lemma (ii) that
\begin{equation*}\lambda(G)\geq \lambda(F_{s,1}) \gt 2+\sqrt{5},\end{equation*} a contradiction. Thereby, 
$G \cong T_{i,j,k}$ with 
$k\geq j\geq i\geq1$. Owe to 
$\lambda(T_{1,2,2})=\frac{5+\sqrt{13}}{2} \gt 2+\sqrt{5}$, then 
$i=j=1$ and 
$k\geq1$, i.e., 
$G \cong T_{1,1,k}$. Consequently, 
$\lambda(T_{1,1,1}) =4$ and 
$\lambda(T_{1,1,k}) \leq \lim_{k \rightarrow \infty}\lambda(T_{1,1,k}) = 2+\sqrt{5}$ by Theorem 3.1.
This finishes the proof.
We finally propose the following problems to end this paper.
Problem 1.
Characterise the connected graphs whose Laplacian matching indices lie in the interval 
$(2+\sqrt{5},\tau]$.
Problem 2.
Determine the limit points of the Laplacian matching indices where these indices are larger than 
$\tau$.
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Conflict of interest
The authors declare that they have no competing interests.
Acknowledgements
Zhaoxi Li and Jianfeng Wang are supported by the National Natural Science Foundation of China (No. 12371353). Jianfeng Wang is supported by the Special Fund for Taishan Scholars Project. Shi-Mei Ma is supported by the National Natural Science Foundation of China (No. 12071063) and the Taishan Scholar Foundation of Shandong Province (No. tsqn202211146).
