Infinite limit of ratio of nth degree polynomials

Here is an explicit formula for your ratio $r_n=\frac{n_n}{d_n}$: $$r_n= \frac{\sum_{k=0}^n\binom{n+k}{2k}(-x)^k} {\sum_{k=0}^n\binom{n+k+1}{2k+1}(-x)^k}.$$ Let $P_n(x)$ and $Q_n(x)$ be the numerator and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.\tag1$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences.

Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$.

In fact (Fedor!), $$r_n=\frac{P_n(x)}{Q_n(x)}=\sqrt{x}\,\frac{U_{2n}(\sqrt{x}/2)}{U_{2n+1}(\sqrt{x}/2)}$$ where $U_n(y)$ are Chebyshev polynomials of the 2nd kind, expressible as $$U_n(y)=\frac{(y+\sqrt{y^2-1})^n-(y-\sqrt{y^2-1})^n}{2\sqrt{y^2-1}}.$$ If $y\geq1$, or equivalently $z=y+\sqrt{y^2-1}\geq1$, then $$\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}= \lim_{n\rightarrow\infty}\frac{z^n-z^{-n}}{z^{n+1}-z^{-n-1}}=\frac1z=y-\sqrt{y^2-1}.$$ If $0<y<1$ then the complex modulus $\vert z\vert=1$ and hence $\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}$ fails to exist.

If $y=0$ then apparently the limit is $0$.

When $y=\frac{\sqrt{x}}2$, the conditions become $$\lim_{n\rightarrow\infty}r_n(x)=\frac{x-\sqrt{x^2-4x}}2$$ if $x\geq4$ or $x=0$. Otherwise (if $0<x<4$) this limit does not exist for being oscillatory!

Finally, since the roots of Chebyshev polynomials $U_n(y)$ lie in $[-1,1]$ it follows that the roots of $U_n(\sqrt{x}/2)$ must be limited in the range $[0,4]$.

The explanation for your observations is that you are dealing with a self-adjoint difference equation in disguise. Let me make a transformation along these line, though I won't analyze it through to the end.

If we let $Y_n=(n_n,d_n)^t$, then this satisfies $$ Y_n = \begin{pmatrix} -1 & x \\ -1 & x-1 \end{pmatrix} Y_{n-1} . $$ Now write $Y_n=A^n W_n$, with $A=\begin{pmatrix} -1 & 0\\ -1&-1\end{pmatrix}$, so $$ A^n = (-1)^n \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} $$ (this is variation of constants, relative to $x=0$). By a calculation, $$ J(W_n-W_{n-1}) = -xH_n W_n, \quad\quad H_n=-JA^{-n}\begin{pmatrix} 0&1\\ 0&1\end{pmatrix}A^{n-1},\quad J=\begin{pmatrix} 0&-1\\ 1&0\end{pmatrix}. $$ An equation of this form is called a (discrete) canonical system, and it corresponds to a symmetric difference expression in the space $\ell^2_H$ if $H_n$ is a positive definite matrix.

Our luck holds here because a calculation gives that $$ H_n = \begin{pmatrix} (n-1)^2 & n-1 \\ n-1 & 1 \end{pmatrix} \ge 0 , $$ as required.

This explains why the zeros are real, though it doesn't show why they are in $[0,4]$. For this, one would have to analyze the spectrum of this system more carefully (you could rewrite it as a Jacobi difference equation).