Finding the limit of $\frac{1}{x}\left(\frac{1}{\arctan x}-\frac{1}{x}\right)$ when $x\to 0$

Question

How to find $$\lim _{x\to 0} \frac{1}{x}\left(\frac{1}{\arctan x}-\frac{1}{x}\right)$$?

I tried doing it by L'Hospital, but I had to do it five times! Is there an easier way around this?

Answer 1

Taylor expansions: much faster, if you know the reference ones. Here, we need the following two: $$\begin{align} \arctan x &= x - \frac{x^3}{3} + o(x^3) \\ \frac{1}{1+x} &= 1-x + o(x) \end{align}$$ (when $x\to 0$). Note that, as a rule of thumb, we take the first one up to the order $3$ as we want to go "further than the first order $x$": since when we'll subtract $\frac{1}{x}$, this will basically "remove the first term of the expansion." We could have gone to higher orders, but that's not necessary: as long as we go far enough, the less expansion the less painful the computations will be.

This leads to $$ \frac{1}{\arctan x} - \frac{1}{x} = \frac{1}{x}\left(\frac{1}{1 - \frac{x^2}{3} + o(x^2)}- 1\right) = \frac{1}{x}\left(1 + \frac{x^2}{3} + o(x^2)- 1\right) = \frac{x}{3} + o(x) $$ so $$ \frac{1}{x}\left(\frac{1}{\arctan x} - \frac{1}{x} \right) = \frac{1}{3} + o(1) \xrightarrow[x\to\infty]{} \frac{1}{3}. $$