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

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?

$$\frac {f(x)} {x^{n}}=x \frac {f(x)} {x^{n+1}}$$. Since $$\frac {f(x)} {x^{n+1}} \to 0$$ and $$x \to 0$$ the product also tends to $$0$$.
If $$f$$ goes to zero faster than $$x^{n+1}$$, then it goes faster than $$x^n$$ ($$x^{n+1}$$ goes faster than $$x^n$$).