# Thread: real zero

1. ## real zero

I was wondering if someone could show me how to do this.
I need to use this fact,
Let p(x) be a polynomial of degree n: p(x) = a0 + a1x + a2x^2 + · · · + anx^n (an cannot equal 0). The limit from x to infinity of p(x)/anx^n is 1 and the limit from x to negative infinity of p(x)/anx^n is 1

to prove that every real polynomial of odd degree has a real zero. I'm not really sure what to do and I would appreciate any help. Thanks in advance.

2. Originally Posted by MKLyon
I was wondering if someone could show me how to do this.
I need to use this fact,
Let p(x) be a polynomial of degree n: p(x) = a0 + a1x + a2x^2 + · · · + anx^n (an cannot equal 0). The limit from x to infinity of p(x)/anx^n is 1 and the limit from x to negative infinity of p(x)/anx^n is 1

to prove that every real polynomial of odd degree has a real zero. I'm not really sure what to do and I would appreciate any help. Thanks in advance.
i think, this has been solved before.. anyways,

$p(x) = a_0 + a_1x + a_2x^2 + \ldots + a_nx^n$ and you want $\lim_{x \rightarrow + \infty} \frac{p(x)}{a_nx^n}$

$\implies \lim_{x \rightarrow + \infty} \frac{p(x)}{a_nx^n} = \lim_{x \rightarrow + \infty} \frac{a_0}{a_nx^n} + \frac{a_1x}{a_nx^n} + \frac{a_2x^2}{a_nx^n} + \ldots + \frac{a_nx^n}{a_nx^n} = 0 + 0 + \ldots + 1 = 1$.. the same way for x approaching neg infty.

now, if n is odd, limit as x approaches positive infinity is 1 while the limit as x approaches negative infinity is -1 (show it) according to a corollary to IVT, if a fuction is continuous on [a,b] and $f(a) \cdot f(b) < 0$, then there is a c on (a,b) st. f(c) = 0..

using this, let a be a very "large" negative number and b be a very large positive number.. then $f(a) \cdot f(b) < 0$, since f(a) is "almost" -1 and f(b) is "almost" 1. hence there is a c in (a,b) s.t f(c) = 0; and this c is a real root (zero).. QED

3. Kalagota: Why divide just take the limit without any division.

4. i think, that is what he wanted to see, and i just explained it that way..