# Limit of a polynomial

1. Nov 15, 2004

### quasar987

Question: How to show that the limit as x goes to infinity of a given polynomial diverge? We have

$$\lim_{x \rightarrow \infty} a_nx^n + a_{n-1}x^{n-1}+...+a_1x+a_0$$

but cannot say the this limit is the sum of the limit of each term separetly because none of these limit exist. We cannot do this either

$$\lim_{x \rightarrow \infty} a_nx^n + a_{n-1}x^{n-1}+...+a_1x+a_0 = \lim_{x \rightarrow \infty} x^n(a_n + \frac{a_{n-1}}{x}+...+\frac{a_1}{x^{n-1}}+\frac{a_0}{x^n})$$

and say the the limit is the product of the limit because one of the limit does not exist. So what then?

Thank you.

2. Nov 16, 2004

### arildno

Let's take the case $$a_{n}>0$$
What you CAN show, using your last equation, is that there exist $$N_{0}$$ so that $$\frac{a_{0}}{x^{n}}\geq{-\frac{a_{n}}{2n}}$$ whenever $$x\geq{N}_{0}$$
Similarly, it exists $$N_{1}$$ so that:
$$\frac{a_{1}}{x^{n-1}}\geq{-\frac{a_{n}}{2n}}$$ whenever $$x\geq{N}_{1}$$
And so on.
Setting N equal to the maximum of these $$N_{i}$$ values, yields the inequality:
$$x^{n}(a_{n}+++\frac{a_{0}}{x^{n}})\geq\frac{x^{n}a_{n}}{2}, x\geq{N}$$