If $p(x)$ is a polynomial then $\lim_{k \to \infty}\frac{p(k+1)}{p(k)}=1$

$$\lim_{k\to\infty}\frac{p(k+1)}{p(k)} =\frac{a_n.(k+1)^n+a_{n-1}(k+1)^{n-1}....a_1}{a_n.k^n+a_{n-1}k^{n-1}....a_{n-1}}$$

$$=\frac{a_n.(1+{1 \over k})^n+{a_{n-1} \over k}(1+ {1 \over k})^{n-1}....{a_1 \over k^n}}{a_n+{a_{n-1} \over k}+....{a_1 \over k^n} }$$ $$=\frac{a_n}{a_n}=1$$

Note that for $x \neq 0$, ${p(x) \over x^n} = a_n + a_{n+1} {1 \over x} + \cdots$.

Hence $\lim_{x \to \infty} {p(x) \over x^n} = a_n$.

Hence $\lim_{x \to \infty} {p(x+1) \over p(x)} = \lim_{x \to \infty}{{p(x+1) \over (x+1)^n} \over {p(x) \over x^n} } = {a_n \over a_n} = 1$.