Limit at Infinity of Polynomial over Complex Exponential

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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:

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


Proof

Let $\epsilon > 0$.

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

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

But:

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


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

\(\displaystyle \frac {\size {\map {P_n} x} } {\size {e^{a x} } }\) \(<\) \(\displaystyle \epsilon\)

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

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


We observe that $M' \ge 0$.

Also, for every $x > M'$:

$x > 0$ as $M' \ge 0$


We have for every $x > M'$:

\(\displaystyle M'\) \(<\) \(\displaystyle x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}\) \(<\) \(\displaystyle x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \frac 1 x \sum_{j \mathop = 0}^n \size {a_j}\) \(<\) \(\displaystyle \epsilon\) 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 $k \in \set {1, \ldots, n + 1}$:

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


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

We get for every $x > M$:

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

$\blacksquare$