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How to show that if $P(x)$ is a polynomial then $\lim_{x \to \infty }P(x) = 0 \iff P(x)=0$,

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marked as duplicate by Nate Eldredge, Ittay Weiss, Asaf Karagila, Micah, rschwieb Feb 24 '13 at 1:51

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Well as i said, use $x^n > x$ for $x>1 $ and $n>1$ –  Dominic Michaelis Feb 23 '13 at 22:15
    
@DominicMichaelis : I know, this is to reinstate the question for the site and everyone else, I wasn't happy to see answers disappeared along with the question. –  Arjang Feb 23 '13 at 22:17
    
@GitGud : I didn't delete the question, I didn't ask the question, it was someone else, I just brought the question back, why are you upset with me? :( –  Arjang Feb 23 '13 at 22:26
    
@Arjang My bad. Sorry. –  Git Gud Feb 23 '13 at 22:26
    
@GitGud : thats cool no worries :) –  Arjang Feb 23 '13 at 22:27

6 Answers 6

up vote 10 down vote accepted

Suppose $P(x) \to 0$ as $x \to \infty$. Let $P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} +\cdots + a_0$. For $x \neq 0$, we can define

$$Q(x) = \frac{P(x)}{x^n} = a_{n} + a_{n-1}x^{-1} + \cdots + a_{0}x^{-n}$$

As $x \to \infty$, $x^{-1},x^{-2}, \cdots, x^{-n}$ all go to zero, so $Q(x) \to a_n$. If $P(x) \to 0$, then we also know that $Q(x) \to 0$. So $a_n = 0$.

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We call the degree of a polynomial $n$.

For $n \ge 1$ $$P(x) = c x^n + O(x^{n-1})$$ (for some nonzero $c$) so limit as $x \to \infty$ is $\pm \infty$ depending on the sign of $c$.

Therefore if $\lim x \to \infty$ gives $0$ then $n = 0$, i.e. $P(x) = 0$.

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@PeterTamaroff en.wikipedia.org/wiki/Proof_by_contradiction –  Gastón Burrull Feb 23 '13 at 23:02
    
too much discussion in comments. I really don't like it. –  user58512 Feb 23 '13 at 23:14
    
@PeterTamaroff $c$ can't be zero, formal negation in this proof is in the degree of the polynomial –  Gastón Burrull Feb 23 '13 at 23:15
    
You're proving that if the polynomial is nonzero, the limit is either infinity or negative infinity, or let's just say nonzero. This is $\neg q \implies \neg p$, which is the same as proving that if the limit is zero, then the polynomial is the zero polynomial. But I don't follow the sentence "Therefore if $\lim$ gives $0$, $n=0$, so $P(x)=0$". I mean, I don't follow why you comment on $n=0$. That is all, your proof is perfectly fine, and in fact it is what I suggested the OP to do. I will delete all comments before this. –  Pedro Tamaroff Feb 23 '13 at 23:20
    
@PeterTamaroff $c=0$ didnt implies that $P=0$ –  Gastón Burrull Feb 23 '13 at 23:23

if $p(x)=0$ it is clear, now for the converse we have$$p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$$ then $$p(x)=x^n(a_n+\frac{a_n-1}{x}+...+\frac{a_0}{x^n})$$ now take the limit as $x\to \infty$ you will get that $$\lim_{x\to \infty}a_nx^n=0$$ since $\lim_{x\to\infty}p(x)=0$ but this is true iff $a_n=0$ ,now you can do the same to get $a_k=0$ where $k\in \{0,...,n-1\}$. Thus, we must have p(x)=0.

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What you obtained is a sieve method to prove it directly. You now obtain $a_n=0$. Continuing the process you get $a_k=0$ for all indices. –  Pedro Tamaroff Feb 23 '13 at 22:23
    
@PeterTamaroff Thats write thank you. –  i.a.m Feb 23 '13 at 22:24
    
Sorry if I confused you. What I meant is that if your hypothesis were "Let $P$ be any polynomial" then using this method (without the assumptions $a_n\neq 0$ or $P\neq 0$ you use to arrive to a contradiction) would yield $P$ is the zero polynomial. –  Pedro Tamaroff Feb 23 '13 at 22:28
    
@PeterTamaroff, so you mean I don't have to assume tht $a_n\ne 0$ –  i.a.m Feb 23 '13 at 22:29
    
Right. See how your method works. I think it is better posed as Isaac did. See how it goes? –  Pedro Tamaroff Feb 23 '13 at 22:33

Call a non-zero polynomial $P(x)$ weird if $\lim_{x\to \infty}P(x)=0$. Suppose there are no weird polynomials of degree $k$. We show there is no weird polynomial of degree $k+1$.

This is easy: if $P(x)$ is weird of degree $k+1$, then $P(x+1)-P(x)$ is weird of degree $k$.

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That $\lim 0=0$ when $x\to \infty$ is immediate. You can prove the converse pretty easily. If $P\neq 0$ then $\lim P\neq 0$ when $x\to\infty$.

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4  
this is just restating the problem, and then saying it's easy. what's the value in that? –  user58512 Feb 23 '13 at 22:38
    
@user58512 Sometimes, proving $\neg q\implies \neg p$ is much easier than proving $p\implies q$. This is an example of this, and leaves the OP some work to do. –  Pedro Tamaroff Feb 23 '13 at 22:58
    
@user58512 OK, you seem to have changed your comment, but what I say stands. If $p$ is not a nonzero polynomial, I think the OP can easily prove that $\lim p\neq 0$. This has a pedagogical value. –  Pedro Tamaroff Feb 23 '13 at 23:07
    
The proof for converse is exactly the same. –  Gastón Burrull Feb 23 '13 at 23:20
    
needless to say that P is not zero to conclude $\lim P/x^n=a_n$ allowing $a_n=a_0=0$ –  Gastón Burrull Feb 23 '13 at 23:29

It is clear that if $P(x)=0$ then the limit as $x$ goes to infinity is $0$.

To prove the converse by contradiction one could do the following. First assume $P(x)$ has positive degree. Therefore the limit as $x$ goes to infinity cannot be $0$ (as $P(x)$ is not bounded) so $P(x)$ must be constant. Finally, it is clear that the only constant polynomial satisfying the required condition is the $0$ polynomial.

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