Question

Show that if f (x) is a polynomial of degree n and g(x) is a polynomial of degree m where m>n, then f (x) is o(g(x)).

Short answer

We have to use the definition of little o notation and prove it is 0.

Step-by-step solution

Step 1:

Let our functions be $\mathrm{f}\left(\mathrm{x}\right)=a_n{x}^n+a_{n-1}{x}^{x^{n-1}}+\dots \dots .+a_0$ and

$\begin{array}{r}g\left(x\right)=b_m{x}^m+b_{m-1}{x}^{m-1}+\dots \dots +b_0\\ \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{a_n{x}^n+a_{n-1}{x}^{x^{n-1}}+\dots \dots +a_0}{m-1^{x^{m-1}}+\dots \dots +b_0}\end{array}$

Step 2:

Dividing numerator and denominator by$x^n$, we get:

$\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{a_n^{x^n}+a_{n-1}{x}^{n-1}+\dots \dots +a_0}{b_m{x}^m+b_{m-1}{x}^{m-1}+\dots \dots +b_0}=0$

Here we notice that f(x) is o(g(x)).