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

This is an approach which mirrors the use of Taylor series.
Let's observe that $$\arctan t =\int_{0}^{t}\frac{dx}{1+x^2}\tag{1}$$ and the expression under limit can be written as $$\left(1+\frac{t^2}{3}\right)\cdot\frac{1}{t^4}\left(\frac{\arctan t} {t} - \frac{1}{1+t^2/3}\right)$$ and hence the limit in question equals the limit of $$\frac{1}{t^4}\left(\frac{\arctan t} {t} - 1+\frac{t^2}{3}+1-\frac{t^2}{3}-\frac{1}{1+t^2/3}\right)$$ The above can be simplified as $$\frac{1}{t^4}\left(\frac{\arctan t} {t} - 1+\frac{t^2}{3}\right) - \frac{1}{9(1+t^2/3)}\tag{2}$$ The first term above can be written as $$\frac{1}{t^5}\int_{0}^{t}\left(\frac{1}{1+x^2}-1+x^2\right)\,dx$$ which is same as $$\frac{1}{t^5}\int_{0}^{t}\frac{x^4}{1+x^2}\,dx$$ Using substitution $x=u^{1/5}$ we can rewrite the above expression as $$\frac{1}{5t^5}\int_{0}^{t^5}\frac{du}{1+u^{2/5}}$$ and this tends to $(1/5)\cdot 1/(1+0^{2/5})=1/5$ via Fundamental Theorem of Calculus.
By equation $(2)$ it is now clear that the desired limit is $1/5-1/9=4/45$.
