little o notation and how we can check it.

Question

I'm working on a problem which asks whether it's possible for $arctan(x) - p(x)$ to be $\textbf{little o}$ of $x^4$ in the limit $x \rightarrow 0$, where $p(x)$ is a third degree polynomial (at most).

Is the way to "check" this then to simply divide by $x^4$ and see if the result is an epsilon function in the given limit?

I attempted this, and if we look at $p(x)/x^4$ then this limit is undefined, and if we look at $arctan(x) / x^4$ and use L'hopital once, we see that this is also undefined. What does this say about the original question?

Answer 1

You want to show $\arctan(x)-p(x)=o(x^4)$ as $x\rightarrow 0$. By definition this means that you want to show:

$$\lim_{x\rightarrow 0}\frac{\arctan(x)-p(x)}{x^4}=0$$

As written, this assertion is immediately false unless you assume $p(0)=0$ (you can't apply l'Hopital's rule otherwise). One way of doing this problem is to write $p(x)=ax^3+bx^2+cx+d$ and then start figuring out what $a,b,c,d$ need to be. We found $d=0$. Apply l'Hopital's rule and you'll figure out another term. Then apply l'Hopital's rule again...

To save you some time, really what you're doing is looking for Taylor expansion of $\arctan(x)$:

$$\arctan(x)=x-x^3/3+x^5/5+o(x^5).$$

So pick $p(x)=x-x^3/3$. Then $\arctan(x)-p(x)=x^5/5+o(x^5)=o(x^4)$.