# How to prove this limit of order up to n?

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!).

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But $(x-a)^{l}\to 0$ as $x\to a$. – Hui Yu Nov 22 '12 at 8:50
You need to make sure that $|x-a|$ is actually less than $1$. One way to do this would be to set $\delta_k = \min\{\delta, 1\}$. – Benjamin Dickman Nov 22 '12 at 8:52
@B.D: Thanks. I really forgot that. – Chan Nov 22 '12 at 8:54

It is a simple application of the product rule which states: If $x_t$ and $y_t$ are two functions such that $\lim_{t \rightarrow a}x_t$ and $\lim_{t \rightarrow a}y_t$ exist, then $$\lim_{t \rightarrow a}x_t y_t = \lim_{t \rightarrow a}x_t \lim_{t \rightarrow a}y_t$$.

But as others have given the rigorous version of the solution, I needn't.

Your proof for the record is indeed correct. While working with multiple limits, consider tactics like taking $\delta < \min(\delta_1,\delta_2)$, where $\delta_1$ and $\delta_2$ are the "$\delta$" for individual sequences(assuming you understand what I am talking about).

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I am very rusty on Maths, so please forgive me for this naive solution. I hope this is not incorrect or incomplete. This is how I see it:

$\displaystyle\lim_{x\to a} \dfrac{f(x)}{(x - a)^n} = 0$

Also,

$\displaystyle\lim_{x\to a} (x - a)^k = 0$ for $k \geq 0$

Also, consider $n - k \geq 0$ if $0 \leq k \leq n$. From this and above,

$\displaystyle\lim_{x\to a} (x - a)^{n - k} = 0$ for $0 \leq k \leq n$

Multiplying both, we get:

$\displaystyle\lim_{x\to a} \dfrac{f(x)}{(x - a)^n} \cdot \displaystyle\lim_{x\to a} (x - a)^{n - k} = 0$ for $0 \leq k \leq n$

$\displaystyle\lim_{x\to a} \dfrac{f(x) \cdot (x - a)^{n - k}}{(x - a)^n} = 0$ for $0 \leq k \leq n$

$\displaystyle\lim_{x\to a} \dfrac{f(x)}{(x - a)^k} = 0$ for $0 \leq k \leq n$

Hence proved.

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