The case $b=0$ is easy, and therefore not considered.

Suppose that $u_n$ is an approximation of $u(x_j)$ with $x_j=j/(n+1))$, where $u(0)=0$ and $u(1)=0$.
A taylor expansion gives
$$
u(x_{j+1})=u(x_{j})+\frac{1}{n+1}u^{\prime}(x_{j})+\frac{1}{2(n+1)^2}u^{(2)}(x_j)+\frac{1}{6(n+1)^3}u^{(3)}(x_j)+\frac{1}{24(n+1)^4}u^{(4)}(\zeta_j)
$$
$$
u(x_{j-1})=u(x_{j})-\frac{1}{n+1}u^{\prime}(x_{j})+\frac{1}{2(n+1)^2}u^{(2)}(x_j)-\frac{1}{6(n+1)^3}u^{(3)}(x_j)+\frac{1}{24(n+1)^4}u^{(4)}(\zeta_j)
$$
Therefore
$$
b u(x_{j+1}) + b u(x_{j-1}) -2 b u(x_{j}) = \frac{b}{(n+1)^2}u^{(2)}(x_j) + O(\frac{1}{n^4}).
$$
$$
b u(x_{j+1}) + b u(x_{j-1}) +j u(x_{j}) = \frac{b}{(n+1)^2}u^{(2)}(x_j) +(2b+ (n+1)x_j) u(x_j)+ O(\frac{1}{n^4}).
$$
So an eigenpair of the matrix is an approximation (for $n$ large) of an eigenpair of
$$
\epsilon^{-2} b u^{\prime\prime} + (2b +\epsilon^{-1} x) u = \lambda u
$$
with $u(0)=u(1)=0$, on $(0,1)$, and $\epsilon=1/(n+1)$.

Multiplying out by $\epsilon^2/b$, and writing $\mu=\epsilon^2\lambda/b$ we find
$$
u^{\prime\prime}+\left(2\epsilon^2 + \frac{\epsilon}{b}x\right)u=\mu u.
$$
So the first order term is negligible for the smallest eigenvalues, (not for the ones of order $\epsilon^{-1}$), and therefore if $b<0$, the first eigenvalues are (up to a mistake in the previous lines)
$$
\lambda_{k}=-\frac{k^2}{n^2}\pi^2 b\left(1+0\left(\frac{1}{n}\right)\right),\mbox{ for } \frac{k}{n} \ll 1.
$$