Limit at Infinity of Polynomial over Complex Exponential

Theorem
Let $n \in \N$.

Let $\map {P_n} x$ be a real polynomial, of degree $n$.

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

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

Then:


 * $\ds \lim_{x \mathop \to +\infty} \frac {\map {P_n} x} {e^{a z} } = 0$

Proof
Let $\epsilon > 0$.

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


 * $\ds x > M \implies \size {\frac {\map {P_n} x} {e^{a z} } - 0} < \epsilon$

But:

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

Let $\ds \map {P_n} x = \sum_{j \mathop = 0}^n \paren {a_j x^j}$ where $a_j \in \R$ for every $j$ in $\set {0, \ldots, n}$.

Let $\ds M' = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}$.

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 $N = \map \max {N_1, \ldots, N_{n + 1} }$.

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

Let $M = \map \max {M', N}$.

We get for every $x > M$: