Prove $\lim_{t\rightarrow 0}\left[-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t\right] = \frac{4}{45}$ The author shows the following limit being taken
$\lim_{t\rightarrow 0}\left[-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t\right] = \frac{4}{45}$
I don't see how you could get anything but $\infty$...? The first term is
$\lim_{t\rightarrow 0}-t^{-4} = \infty$...
Or, finding a common denominator:
$-t^{-4} + t^{-5}\left(1+\frac{t^2}{3}\right)\tan^{-1}t = \frac{t^5 + t^4(1+\frac{t^2}{3})\tan^{-1}t}{t^9} = \frac{3t^5 + (3t^4 + t^6)\tan^{-1}t}{3t^9}$
which doesn't illuminate anything

With little $o$ notation,$$\begin{align}\frac{-t+(1+\tfrac13t^2)\arctan t}{t^5}&=\frac{-1+(1+\tfrac13t^2)(1-\tfrac13t^2+\tfrac15t^4+o(t^4))}{t^4}\\&=\frac{-\tfrac19t^4+\tfrac15t^4+o(t^4)}{t^4}\\&=-\tfrac19+\tfrac15+o(1)\\&=\tfrac{4}{45}+o(1).\end{align}$$
