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
 A: With little $o$ notation,$$\begin{align}\frac{-t+(1+\tfrac13t^2)\arctan t}{t^5}&=\frac{-1+(1+\tfrac13t^2)(1-\tfrac13t^2+\tfrac15t^4+o(t^4))}{t^4}\\&=\frac{-\tfrac19t^4+\tfrac15t^4+o(t^4)}{t^4}\\&=-\tfrac19+\tfrac15+o(1)\\&=\tfrac{4}{45}+o(1).\end{align}$$
A: 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. 
A: Hint. Factor out $t^{-5},$ to get $$\frac{1}{t^5}\left(-t+\left(1+\frac{t^2}{3}\right)\arctan t\right).$$ Then use L'hopital.
A: 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 $1/5-1/9=4/45$.
