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

Let $$p(x)$$ be a polynomial of degree $$n$$, i.e $$p(x)=a_nx^n+a_{n-1}x^{n-1}+...a_0, a_n\neq0$$

For every $$\varepsilon >0$$, we have to find a stage such that after that stage it should be like $${\left|\frac{p(k+1)}{p(k)}-1\right|<\varepsilon},$$

i.e. $${\left|\frac{p(k+1)-p(k)}{p(k)}\right|=\frac{|a_n((k+1)^n-k^n))+...a_2((k+1)^2-k^2)+a_1|}{|a_nk^n+a_{n-1}k^{n-1}+...a_0|}}$$

I dont know how to proceed further

• The limit is true for any nonzero polynomial. Divide the numerator and denominator by $a_nx^n$... – Jean-Claude Arbaut Sep 6 at 15:09
• The term $(k+1)^n-k^n$ does not contain any term of order $n$, so... – Mostafa Ayaz Sep 6 at 15:09
• Is n't it obvious if you divide $\frac{p(k)+1}{p(k)}$ by $k^n$ – Rishi Sep 6 at 15:39
• Do you wish a numerical approximation in terms of $a_i$ and $\epsilon$ for the first $K$ such that for all $k \geq K$ this holds true? – clark Sep 6 at 16:32

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