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.

\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 $x^2$. 
