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While what I'm trying to prove is related to this topic but alas I cannot just assume that given: $ p(x) $ where $ \deg(p(x))=2n+1$ where $n\in\mathbb{N}$ that the only two cases are:

  • $\lim \limits_{x\to -\infty}f(x)=-\infty$ and $\lim \limits_{x\to \infty}f(x)=\infty$

or

  • $\lim \limits_{x\to -\infty}f(x)=\infty$ and $\lim \limits_{x\to \infty}f(x)=-\infty$

I've tried proving that given that

$ x^{2n+1} $ and $ \sum \limits_{k=0}^{2n}{a_{k}x^k}$, $n\in\mathbb{N}$ and $a_k>1$

that

$ \exists x_1\in\mathbb{R} , \forall x>x_1 $

$ x^{2n+1} > \sum \limits_{k=0}^{2n}{a_{k}x^k} $

And I think I proved it but it's an ugly inductive proof: for $n=1$

$ x^{3} > a_2x^2+a_1x+a_0 $

then I've tried choosing $x_1 = (a_2a_1a_0)^2 $

then assuming $ x^{2n+1} > \sum \limits_{k=0}^{2n}{a_{k}x^k} $ proving that $ x^{2n+3} > \sum \limits_{k=0}^{2n+2}{a_{k}x^k} $

Which is the same as : $ x^{2n+3} > \sum \limits_{k=0}^{2n}{a_{k}x^k} + a_{2n+1}x^{2n+1}+ a_{2n+2}x^{2n+2} $ and then replace $ \sum \limits_{k=0}^{2n}{a_{k}x^k} $ with $ x^{2n+1} $

meaning : $ x^{2n+3} > x^{2n+1} + a_{2n+1}x^{2n+1}+ a_{2n+2}x^{2n+2} $ and assuming $x\neq0$ dividing by $ x^{2n+1} $ will mean that it's sufficient to show that:

$x^{2} > 1 + a_{2n+1} + a_{2n+2}x $ which is too fairly easy to show.

The problem is I don't know whether showing this is really sufficient ( I intended to use that to show that if for example the coefficient of the highest degree variable is positive that $\forall M\in\mathbb{R} \exists X_1 , x>X_1 \Rightarrow f(x)>M$ ).

So what I'm asking is:

  1. Whether it makes sense at all because it feels like I have holes in my proof.
  2. Is there a more mathematically elegant way to prove this not using derivatives.

Thanks and I deeply apologize if I broke any rule, I've really tried not to.

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1 Answer 1

up vote 3 down vote accepted

For an elegant approach, the usual method for polynomials and rational functions appearing in limits is to factor out the most significant term. e.g.

$$ \lim_{x \to +\infty} 1 + 2x + x^2 = \lim_{x \to +\infty} x^2 \left(\frac{1}{x^2} + \frac{2}{x} + 1 \right)$$

$$ \lim_{x \to 1} 1 - x - x^2 + x^3 = \lim_{x \to 1} (x-1)^2 \left((x-1) + 2 \right) $$

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So you propose to write that given the previously mentioned $p(x)$ I can always describe it as $ a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + \cdots + a_{2}x^{2} + a_{0}$ and then just factor it into $ \lim\limits_{x \to +\infty} x^{2n+1} \left(\frac{a_0}{x^{2n+1}} + \cdots + \frac{a_{2n}}{x} + a_{2n+1} \right) $ ? And then using limit properties show that this limit is indeed $\pm\infty$ depending on the sign of $a_{2n+1}$ when $x\to\pm\infty$ ? –  Scis Jan 12 '13 at 9:50
    
Right, that's the general idea. –  Hurkyl Jan 12 '13 at 11:07

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