Finding the $\lim\limits_{x \to 0+} (\frac{1}{x} - \frac{1}{\arctan x})$ Right so, so far I have gotten the denominators the same and have the achieved the indeterminate 0/0 as $\lim\limits_{x \to 0+} \frac{\arctan x - x}{x\arctan x}$ once I apply L' Hospital's Rule though, it gets really messy. So I am wondering if there is a more elegant or efficient way to find the answer without having to deal with the nesting fractions.
 A: \begin{align}
& \lim_{x\to0} \frac{\arctan x - x}{x\arctan x} = \lim_{x\to0} \frac{\left(\dfrac 1 {1+x^2} \right) - 1}{x\left( \dfrac 1 {1+x^2} \right) + 1\cdot\arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac{1 - (1+x^2)}{ x + (1+x^2) \arctan x } = \lim_{x\to0} \frac{-x^2}{x + (1+x^2)\arctan x} \\[10pt]
= {} & \left( \lim_{x\to0} \frac {\left(\dfrac x {\arctan x}\right)} {\left( \dfrac x {\arctan x} \right) + (1+x^2)} \right) \cdot \left( \lim_{x\to0} (-x) \right) \\[10pt]
\end{align}
Now use the fact that $\displaystyle\lim_{x\to0} \frac x {\arctan x} = 1.$
This is to be expected because $x \arctan x$ crosses the axis with a slope of $1$ and is an odd function, so no $x^2$ term can appear in its power series, so in $\arctan x - x$ the $x$ term cancels and there's no $x^2$ term, so it's like $x^3 + \text{higher-degree terms}$, whereas $x\arctan x$ is like $x^2$. 
A: With $y=\arctan(x)$, we get
$$\lim_{x\to 0^+}\frac{\arctan(x)-x}{x\arctan(x)}=\lim_{y\to 0^+}\frac{y-\tan(y)}{\tan(y)y}$$
$$=\lim_{y\to 0^+}\frac{y-\tan(y)}{y^2}\frac{y}{\tan(y)}$$
$$=\lim_{y\to 0^+}\frac{1-(1+y^2)}{2y}*1$$
$$=\lim_{y\to 0^+}\frac{-y^2}{2y}=0$$
A: May be, you could have used Taylor series $$\tan^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ So $$\frac 1{\tan^{-1}(x)}=\frac 1{x-\frac{x^3}{3}+O\left(x^4\right)}=\frac 1 x\times \frac 1{1-\frac{x^2}{3}+O\left(x^3\right)}=\frac 1 x\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)$$ which finally make $$\frac 1 x-\frac 1{\tan^{-1}(x)}=\frac 1 x-\frac 1 x\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)=-\frac{x}{3}+O\left(x^2\right)$$
