Limit at Infinity of Polynomial over Complex Exponential

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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 function, 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:

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


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

\(\displaystyle \frac {\left \vert {P_n\left({x}\right)} \right \vert}{\left \vert {e^{a x} } \right \vert}\) \(<\) \(\displaystyle \epsilon\)

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 $x > M'$:

\(\displaystyle M'\) \(<\) \(\displaystyle x\)
\(\displaystyle \iff \ \ \) \(\displaystyle \frac 1 \epsilon \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert\) \(<\) \(\displaystyle x\)
\(\displaystyle \iff \ \ \) \(\displaystyle \frac 1 x \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert\) \(<\) \(\displaystyle \epsilon\) 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 $N = \max \left({N_1, \ldots, N_{n + 1}}\right)$.

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

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


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

We get for every $x > M$:

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

$\blacksquare$