Show that for $ n \in N $, if a function $f(x)$ is $o(x^{n+1})$, then $f(x)$ is $o(x^n)$

Question

By $o(x^n)$, I mean the little-o of $x^n$, not to be confused with the big-O notation.

The definition of $o(x^n)$ in my book is: In general, for any natural number n, if a function $f(x)$ satisfies $f(0)=0$ and $lim_{x\to 0} {f(x)\over x^n} = 0$, then we know that the function $f(x)$ converges to $0$ near the origin much faster then $x^n$ converges to $0$. Such a function $f$ is denoted as $f(x) \in o(x^n)$

The way I tried to solve this problem is by drawing the graphs of $x^n$ when $n$ is $1$ to $6$ by a graphing calculator. As $n$ increases, the graph of $x^n$ was converging to $0$ when it's near the origin faster and faster. Therefore, my conclusion was that if $f(x)$ converges to $0$ near the origin much faster then $x^{n+1}$, $f(x)$ has to converge to $0$ faster than $x^n$.

Is this way of thinking legit? And is there a better way to approach this problem?

Answer 1

If $f$ goes to zero faster than $x^{n+1}$, then it goes faster than $x^n$ ($x^{n+1}$ goes faster than $x^n$).

This sort of thing is "transitive".

Answer 2

$\frac {f(x)} {x^{n}}=x \frac {f(x)} {x^{n+1}}$. Since $\frac {f(x)} {x^{n+1}} \to 0$ and $x \to 0$ the product also tends to $0$.