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

As suggested, consider the Taylor expansion of $\arctan t =  \frac{t^3}{3} + \frac{t^5}{5} - \frac{t^7}{7} + \cdots$
If you use this and ignore the terms with positive powers of $t$ (they all tend to zero), you should get your answer. I'm interested to see if there are other ways around it aside from Taylor polynomials. 
