Limit development for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. I'm trying to solve this problem: Find a limit development of order
7 for the function $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. Where
we use the next definition:

Definition. Let $f\colon I\to\mathbb{R}$ be a function and $x_{0}\in I$. We
  say that $f$ has a limit development of order $n$ in $x_{0}$ provided
  that there exist $a_{0},a_{1},\ldots,a_{n}\in\mathbb{R}$ such that
  for $x\in I$
$$
f(x)=a_{0}+a_{1}(x-x_{0})+\ldots+a_{n}(x-x_{0})^{n}+o((x-x_{0})^{n}) \text{ (small $o$)}.
$$

Using Taylor series for $\arctan x$ at 0, we have that $\arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots$,
and therefore 
\begin{align*}
\arctan(x^{2}) & =x^{2}-\frac{(x^{2})^{3}}{3}+\frac{(x^{2})^{5}}{5}-\frac{(x^{2})^{7}}{7}+\ldots\\
 & =x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots
\end{align*}
So, replacing for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$ I have
that:
\begin{align*}
f(x) & =\frac{x}{x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots}-\frac{1}{x}\\
 & =\frac{x}{x\left(x-\frac{x^{5}}{3}+\frac{x^{9}}{5}-\frac{x^{13}}{7}+\ldots\right)}-\frac{1}{x}\\
 & =\frac{1}{x}\cdot\left(1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots\right)^{-1}-\frac{1}{x}
\end{align*}
But I couldn't obtain the form that is necessary for the limit development
because I have that part with the inverse of $1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots$.
Could you help me or give me some suggestion?
Thanks.
 A: We may start with few terms of the Maclaurin series of $\frac{\tan x}{x}$ (it is not that painful to compute a few derivatives of the tangent function at the origin)
$$ \frac{\tan x}{x}=1+\frac{x^2}{3}+\frac{2 x^4}{15}+O(x^6)$$
then replace $x$ with $\arctan(z)=z-\frac{z^3}{3}+\frac{z^5}{5}+o(z^6)$, giving
$$ \frac{z}{\arctan z} = 1+\frac{z^2}{3}-\frac{4z^4}{45}+O(z^6)$$
(this is related to Gregory coefficients, by the way) and, by setting $z=t^2$,
$$ \frac{t^2}{\arctan(t^2)} = 1+\frac{t^4}{3}-\frac{4t^8}{45}+O(t^{12})$$
$$ \frac{t}{\arctan(t^2)} = \frac{1}{t}+\frac{t^3}{3}-\frac{4t^7}{45}+O(t^{11})$$
then finally
$$ \frac{x}{\arctan(x^2)}-\frac{1}{x} = \frac{x^3}{3}-\frac{4x^7}{45}+O(x^{11}).$$
A: You could use the taylor polynomial of $(1-x)^{-1}$ as well, because:
$$\frac{1}{1-x}=\sum_{n \in \mathbb{N}} x^n$$
A: $$f(x)=\dfrac{x^2-\arctan(x^2)}{x^6}\cdot\dfrac{x^6}{x\arctan(x^2)}$$
Now use series expansion for $\arctan(x^2)$ for the first ratio or use the last problem of Are all limits solvable without L'Hôpital Rule or Series Expansion
Can you take up the second?
A: Just to check people's answers,
Wolfy says
$x^3/3 - (4 x^7)/45 + (44 x^{11})/945 + O(x^{13})
$.
