$P(x) \sim_{ n \rightarrow \infty} x^{n}$

Question

If $P$ is a monic polynomial and $x$ is a real number $>1$.

So $P(x) = x^{n} + a_{n-1} \cdot x^{n-1} +... + a_0$

In this case do we have : $P(x) \sim_{n \rightarrow \infty} x^{n}$ ?

My first instinct was to factorize : $P(x) = x^{n} \cdot (1 + \text{ some fractions})$, and because $ x > 1$ we have : $\lim_{n \rightarrow \infty} x^{n} \cdot (1 + \text{ some fractions}) = \lim_{n \rightarrow \infty} x^n$

But I don't know if this is correct...

Answer 1

Best way to check is to evaluate the limit $$\lim_{n \to \infty} \frac{P(x)}{x^n}=a$$

If $a=1$, then the sign $\sim$ is used.

In your question, the limit isn't necessarily $1$, instead it may be finitely greater than $1$.It may even not exist at all, depending on the coefficients.

So, according to the given Information, we can't say anything for certain between the asymptotic relation of the polynomial and $x^n$.

Answer 2

If $x>1$ and $0\le i<j$, $ x^i=o(x^j)$, so for any polynomial of degree $n$ $\;p(x)=a_n x^n+ \text{terms of lower degree}$, we have $$p(x)=a_n\mkern1mu x^n+o(x^n),\enspace\text{so by definition}\quad p(x)\sim_{x\to\infty}a_n \mkern1mu x^n.$$