Let $P$ a polynomial with positive coefficients.Determine $\lim_{ x\to\infty }\frac{ \lfloor P(x) \rfloor}{P(\lfloor x \rfloor)}$

Question

this is my idea. By definition $p(x)-1\leq \left \lfloor p(x) \right \rfloor \leq p(x)$ and $x-1 \leq \left \lfloor x \right \rfloor \leq x$. As $p(x)$ is increasing and positive in $(0, \infty )$ then $\frac{1}{p(x)}\leq \frac{1}{p(\left \lfloor x \right \rfloor)}\leq \frac{1}{p(x-1)}$,this is $1-\frac{1}{p(x)}\leq \frac{\left \lfloor p(x) \right \rfloor}{p(\left \lfloor x \right \rfloor)}\leq \frac{p(x)}{p(x-1)}$. I want to prove that $\lim_{ x\to\infty }\frac{ \lfloor P(x) \rfloor}{P(\left \lfloor x \right \rfloor)}=1$. this is correct?. Help me please!!

Answer 1

Hint If nonconstant, the polynomials $p(x-1), p(x),$ and $p(x)-1$ all have the same leading term.