Given that $\displaystyle\lim_{x\to a} \dfrac{f(x)}{(x - a)^n} = 0$ Prove that for any given $k, 0 \leq k \leq n$, then $\displaystyle\lim_{x\to a} \dfrac{f(x)}{(x - a)^k} = 0$
Here is my current work,
Choose an arbitrary $\epsilon > 0$, choose $\delta = 1$. From hypothesis, we have that if $|x - a| < \delta \Rightarrow \bigg|\dfrac{f(x)}{(x - a)^n} - 0\bigg| = \bigg|\dfrac{f(x)}{(x - a)^n}\bigg| < \epsilon$ It follows that if $|x - a| < \delta$, then $\bigg|\dfrac{f(x)}{(x - a)^k}\bigg| = \bigg|\dfrac{f(x)}{(x - a)^n}\bigg| \cdot |(x - a)^l| < \epsilon \cdot 1^{l} = \epsilon$
Does this proof make sense? Any suggestion would be greatly appreciated.
In fact, my initial approach is to consider $\bigg|\dfrac{f(x)}{(x - a)^n}\bigg| \text{ vs. } \bigg|\dfrac{f(x)}{(x - a)^k}\bigg|$ but I couldn't deduce from that to be less than $\epsilon > 0$ because $k \leq n$, so $|(x - a)^n|$ could be either larger or smaller than $|(x - a)^n|$ depends on $x$ and $a$. (Well actually while writing this question, I've just realized that as long $|x - a| < 1$, it works!).