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.

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 $$\frac 1 x-\frac 1{\tan^{-1}(x)}=\frac 1 x-\frac 1 x\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)=-\frac{x}{3}+O\left(x^2\right)$$
