= Solution
Use the <characteristic polynomial> in the convention $p(\lambda)=\det(\lambda I-A)$:
$$
p(\lambda)=(\lambda+1)(\lambda-\alpha)(\lambda-1)-\epsilon(3\lambda-1).
$$
For fixed $\alpha>0$ with $\alpha\ne1$, all three limiting <eigenvalues> are simple. Expanding this <polynomial> at each root, or applying <first-order perturbation of a simple eigenvalue>, gives
$$
\boxed{\begin{aligned}
\lambda_{-1}&=-1-\frac{2\epsilon}{\alpha+1}+O(\epsilon^2),\\
\lambda_\alpha&=\alpha+\frac{(3\alpha-1)\epsilon}{(\alpha+1)(\alpha-1)}+O(\epsilon^2),\\
\lambda_1&=1-\frac{\epsilon}{\alpha-1}+O(\epsilon^2).
\end{aligned}}
$$
These are fixed-$\alpha$ <asymptotic expansions>, not uniform through $\alpha=1$. At $\alpha=1/3$ the displayed correction to $\lambda_\alpha$ vanishes because $\lambda=1/3$ is an exact root for every $\epsilon$; there is no second nonzero term for that branch.
The two roots near $1$ interact when the correction $\epsilon/(\alpha-1)$ is comparable to their unperturbed gap $\alpha-1$. Thus the <distinguished limit> is $\alpha=1+d\sqrt\epsilon$, with bounded real $d$. Set $h=\sqrt\epsilon$ and $\lambda=1+hL+h^2N+\cdots$. The <polynomial> becomes
$$
(2+\delta)\delta(\delta-dh)-h^2(2+3\delta)=0,\qquad \delta=\lambda-1.
$$
At orders $h^2$ and $h^3$,
$$
L(L-d)=1,\qquad N(2L-d)=L.
$$
With $D=\sqrt{d^2+4}$, the two interacting <eigenvalues> are
$$
\boxed{\lambda_\pm=1+\frac{h}{2}(d\pm D)+\frac{h^2}{2}\left(1\pm\frac dD\right)+O(h^3),\qquad \lambda_{-1}=-1-\epsilon+O(\epsilon^{3/2}).}
$$
The $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha=1$ is $d=0$:
$$
\boxed{\lambda_\pm=1\pm\sqrt\epsilon+\frac\epsilon2+O(\epsilon^{3/2}),\qquad \lambda_{-1}=-1-\epsilon+O(\epsilon^2).}
$$
The half-integer powers reflect <square-root splitting of a defective double eigenvalue>: the limiting <matrix> has a nontrivial <Jordan block> at $1$.
For matching, write $\Delta=\alpha-1$ and take the overlap $\sqrt\epsilon\ll|\Delta|\ll1$. For $d\to+\infty$, $L_+=d+d^{-1}+O(d^{-3})$, $N_+=1+O(d^{-2})$, while $L_-=-d^{-1}+O(d^{-3})$, $N_-=O(d^{-2})$. Consequently the distinguished branches reduce to
$$
\lambda_{\alpha}=\alpha+\frac\epsilon\Delta+\epsilon+O(\epsilon\Delta)+O(\epsilon^2/|\Delta|^3),\qquad
\lambda_1=1-\frac\epsilon\Delta+O(\epsilon^2/|\Delta|^3).
$$
The regular coefficient satisfies $(3\alpha-1)/[(\alpha+1)(\alpha-1)]=1/\Delta+2/(2+\Delta)=1/\Delta+1+O(\Delta)$, proving agreement. For $d\to-\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha+1)=-1+O(\Delta)$. As a spectral check, <positive diagonal symmetrization of a matrix> transforms this tridiagonal <matrix> to a real <symmetric matrix> with off-diagonal entries $\sqrt{2\epsilon}$ and $\sqrt\epsilon$; its <eigenvalues> are real for positive $\epsilon$.
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