Theorem: Limit at Infinity of Polynomial over Complex Exponential

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:
{{begin-eqn}}
{{eqn | l = \size {\frac {\map {P_n} x}{e^{a z} } - 0}
      | r = \frac {\size {\map {P_n} x} } {\size {e^{a z} } }
      | c = Modulus of Product
}}
{{eqn | r = \frac {\size {\map {P_n} x} } {\size {e^{a x \mathop + i a y} } }
}}
{{eqn | r = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }
      | c = Modulus of Exponential is Modulus of Real Part
}}
{{end-eqn}}
This means it is sufficient to find an $M \in \R$ such that:
{{begin-eqn}}
{{eqn | l = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }
      | o = <
      | r = \epsilon
}}
{{end-eqn}}
{{questionable|Exponential Dominates Polynomial doesn't say anything about the quotient. (Proof corrected Nov. 2016.)}}
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 $x > M'$:
{{begin-eqn}}
{{eqn | l = M'
      | o = <
      | r = x
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}
      | o = <
      | r = x
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac 1 x \sum_{j \mathop = 0}^n \size {a_j}
      | o = <
      | r = \epsilon
      | c = as $x \not = 0$ since $x > 0$
}}
{{end-eqn}}
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 $k \in \set {1, \ldots, n + 1}$:
{{begin-eqn}}
{{eqn | l = x^k
      | o = <
      | r = e^{a x}
      | c = for all $x > N$
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {x^k} {e^{a x} }
      | o = <
      | r = 1
      | c = for all $x > N$
}}
{{end-eqn}}
Let $M = \map \max {M', N}$.
We get for every $x > M$:
{{begin-eqn}}
{{eqn | l = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }
      | r = \frac {\size {\sum_{j \mathop = 0}^n \paren {a_j x^j} } } {e^{a x} }
}}
{{eqn | o = \le
      | r = \frac {\sum_{j \mathop = 0}^n \size {a_j x^j} } {e^{a x} }
}}
{{eqn | r = \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j}  x^j} } {e^{a x} }
      | c = as $x^j > 0$ since $x > 0$
}}
{{eqn | r = \frac 1 x \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j} x^\paren {j + 1} } } {e^{a x} }
      | c = as $x \not = 0$ since $x > 0$
}}
{{eqn | r = \frac 1 x \sum_{j \mathop = 0}^n \paren {\size {a_j} \frac {x^\paren {j + 1} } {e^{a x} } }
}}
{{eqn | o = \le
      | r = \frac 1 x \sum_{j \mathop = 0}^n \size {a_j} 1
      | c = as $\dfrac {x^\paren {j + 1} } {e^{a x} } < 1$ for every $j$ in $\set {0, \ldots, n}$
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
{{qed}}
Category:Limits of Mappings
Category:Polynomial Theory
Category:Exponential Function
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