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:

 $\ds \size {\frac {\map {P_n} x}{e^{a z} } - 0}$ $=$ $\ds \frac {\size {\map {P_n} x} } {\size {e^{a z} } }$ Modulus of Product $\ds$ $=$ $\ds \frac {\size {\map {P_n} x} } {\size {e^{a x \mathop + i a y} } }$ $\ds$ $=$ $\ds \frac {\size {\map {P_n} x} } {\size {e^{a x} } }$ Modulus of Exponential is Modulus of Real Part

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

 $\ds \frac {\size {\map {P_n} x} } {\size {e^{a x} } }$ $<$ $\ds \epsilon$

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 $x > M'$:

 $\ds M'$ $<$ $\ds x$ $\ds \leadstoandfrom \ \$ $\ds \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}$ $<$ $\ds x$ $\ds \leadstoandfrom \ \$ $\ds \frac 1 x \sum_{j \mathop = 0}^n \size {a_j}$ $<$ $\ds \epsilon$ as $x \not = 0$ since $x > 0$

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 $k \in \set {1, \ldots, n + 1}$:

 $\ds x^k$ $<$ $\ds e^{a x}$ for all $x > N$ $\ds \leadstoandfrom \ \$ $\ds \frac {x^k} {e^{a x} }$ $<$ $\ds 1$ for all $x > N$

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

We get for every $x > M$:

 $\ds \frac {\size {\map {P_n} x} } {\size {e^{a x} } }$ $=$ $\ds \frac {\size {\sum_{j \mathop = 0}^n \paren {a_j x^j} } } {e^{a x} }$ $\ds$ $\le$ $\ds \frac {\sum_{j \mathop = 0}^n \size {a_j x^j} } {e^{a x} }$ $\ds$ $=$ $\ds \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j} x^j} } {e^{a x} }$ as $x^j > 0$ since $x > 0$ $\ds$ $=$ $\ds \frac 1 x \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j} x^\paren {j + 1} } } {e^{a x} }$ as $x \not = 0$ since $x > 0$ $\ds$ $=$ $\ds \frac 1 x \sum_{j \mathop = 0}^n \paren {\size {a_j} \frac {x^\paren {j + 1} } {e^{a x} } }$ $\ds$ $\le$ $\ds \frac 1 x \sum_{j \mathop = 0}^n \size {a_j} 1$ as $\dfrac {x^\paren {j + 1} } {e^{a x} } < 1$ for every $j$ in $\set {0, \ldots, n}$ $\ds$ $<$ $\ds \epsilon$

$\blacksquare$