Let me try to expand a little bit the problem (so it's too long for a usual comment).

Replace $a_i/b_i$ and $c_i/b_i$ with $a_i$ and $c_i$, respectively (for obvious simplification reasons), and consider the determinant $D_N=D_N(\lambda;a_1,\dots,a_N;c_0,\dots,c_{N-1})$ of the corresponding matrix $\lambda-A$. Expanding the determinant along the first row gives $$ D_N(\lambda;a_1,\dots,a_N;c_0,\dots,c_{N-1}) =\lambda D_{N-1}(\lambda;a_2,\dots,a_N;c_1,\dots,c_{N-1}) -c_0a_1D_{N-2}(\lambda;a_3,\dots,a_N;c_2,\dots,c_{N-1}); $$ in other words, $$ D_N/D_{N-1}=\lambda-\frac{a_0c_1}{D_{N-1}/D_{N-2}} =\dots =\lambda-\frac{a_0c_1}{\lambda-\dfrac{a_1c_2}{\lambda-\dfrac{a_2c_3}{\dots -\dfrac{a_Nc_{N-1}}{\lambda}}}}. $$ In order to get some information about the asymptotics of the zero(s) of $D_N(\lambda)/D_{N-1}(\lambda)$ one really have to have some knowledge about the $a_ic_{i-1}$, $i=1,2,\dots$. This reduces the problem to a problem of the related family of orthogonal polynomial and even Deift's book is too advanced, it is the best source on this.

Let me try to expand a little bit the problem (so it's too long for a usual comment).

Consider the determinant $D_N=D_N(\lambda;a_1,\dots,a_{N-1};b_1,\dots,c_{N-1})$ of the corresponding matrix $\lambda-A$. Expanding the determinant along the first row gives $$ D_N(\lambda;a_1,\dots,a_{N-1};b_1,\dots,b_{N-1}) =\lambda D_{N-1}(\lambda;a_2,\dots,a_{N-1};b_2,\dots,b_{N-1}) -a_1b_1D_{N-2}(\lambda;a_3,\dots,a_{N-1};b_3,\dots,b_{N-1}); $$ in other words, $$ D_N/D_{N-1}=\lambda-\frac{a_1b_1}{D_{N-1}/D_{N-2}} =\dots =\lambda-\frac{a_1b_1}{\lambda-\dfrac{a_2b_2}{\lambda-\dfrac{a_3b_3}{\dots -\dfrac{a_{N-1}b_{N-1}}{\lambda}}}}. $$ In order to get some information about the asymptotics of the zero(s) of $D_N(\lambda)/D_{N-1}(\lambda)$ one really have to have some knowledge about the $a_ib_i$, $i=1,2,\dots$. This reduces the problem to a problem for the related family of orthogonal polynomials and even Deift's book is too advanced, it is the best source on this.