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How could I formalize that, when a polinomial $P(x)=a_0+a_1 x +\ldots + a_n x^n$ satisfies $$ \lim_{x\to\infty}P(x) = 0 $$ then $P(x)=0$ for all $x\in\mathbb R$? I've tried to show that supposing that for all $\varepsilon > 0$ there exists $K > 0$ so that $K < x$ then $|P(x)| < \varepsilon$ implies that $a_0 = ... = a_n = 0$ using reductio ad absurdum and other ways but did not get any relevant results. Thanks in advance

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  • $\begingroup$ Show $x^n > x $ for $x>1$ $\endgroup$ – Dominic Michaelis Feb 23 '13 at 21:37
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    $\begingroup$ Please do not delete questions with good answers. Others have devoted effort to answer your question; deleting the question is disrespectful of their effort and prevents others from benefiting from your question and its answers. $\endgroup$ – robjohn Mar 8 '13 at 11:06
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Prove the contrapositive in a simple way:

Suppose $P(x)\neq 0$. Then there's some $i\in \{0, \dots n\}$ such that $a_i\neq 0$.

Therefore $C:=\{i\in \{0, \dots n\} : a_i\neq 0\}\neq \varnothing$ and it follows that $C$ has a maximum $k$.

  1. If $k=0$, then $P(x)=a_0\neq 0$ and it follows that $\displaystyle \lim _{x\to +\infty}\left(P(x)\right)=a_0\neq 0$.
  2. If $k\neq 0$, then $\displaystyle \lim _{x\to +\infty} \left(P(x)\right)=\lim _{x\to +\infty}\left(a_kx^k\right)=\pm \infty$ and $\displaystyle \lim _{x\to +\infty}\left(P(x)\right)\neq 0$.

The contrapositive was proved.

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