Limit at Infinity of Polynomial over Complex Exponential

Theorem
Let $n \in \N$.

Let $P_n \left({x}\right)$ be a real polynomial, of degree $n$.

Let $e^z$ be the complex exponential, where $z = x + i y$.

Let $a \in \R_{>0}$.

Then:


 * $\displaystyle \lim_{x \mathop \to +\infty} \frac {P_n \left({x}\right)}{e^{a z}} = 0$

Proof
Let $\epsilon > 0$.

By the definition of limits at infinity, we need to show that there is some $M \in \R$ such that:


 * $\displaystyle x > M \implies \left \vert {\frac {P_n\left({x}\right)}{e^{a z}} - 0} \right\vert < \epsilon$

But:

This means it is sufficient to find an $M \in \R$ such that:

Let $\displaystyle P_n \left({x}\right) = \sum_{j \mathop = 0}^n \left({a_j x^j}\right)$ where $a_j \in \R$ for every $j$ in $\left\{{0, \ldots, n}\right\}$.

Let $\displaystyle M' = \frac 1 \epsilon \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert$.

We observe that $M' \ge 0$.

Also, for every $x > M'$:
 * $x > 0$ as $M' \ge 0$

We have for every $x > M'$:

We have that for every $k \in \N$ there exists $N_k \in \N$ such that:
 * $x^k < e^{a x}$ for all $x > N_k$ by Exponential Dominates Polynomial

Let $N = \max \left({N_1, \ldots, N_{n + 1}}\right)$.

We have for every $k \in \left\{{1, \ldots, n + 1}\right\}$:

Let $M = \max \left({M', N}\right)$.

We get for every $x > M$: