# 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

• Try taking a common denominator. Commented Apr 18, 2020 at 19:21
• @Tavish thanks, but I don't see anything useful in doing that Commented Apr 18, 2020 at 19:35
• What is the degree-3 Taylor polynomial of $\tan^{-1}(t)$? And you're not doing a common denominator in a reasonable way. Commented Apr 18, 2020 at 19:36
• @TedShifrin Am I? What is a reasonable way? Commented Apr 18, 2020 at 19:39
• Actually, $\lim_{t\to0}-t^{-4}=-\infty$. But just because the terms have limits $\pm\infty$, doesn't mean their sum doesn't converge.. See also here.
– J.G.
Commented Apr 18, 2020 at 19:40

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}

• This is the most straightforward answer, thanks Commented Apr 18, 2020 at 20:14

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.

• yes this works, thanks! Commented Apr 18, 2020 at 20:14

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.

• This is what I meant. Commented Apr 18, 2020 at 19:38
• Taylor polynomials are infinitely better for such things than L'Hôpital. Commented Apr 18, 2020 at 19:41

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 $$1/5-1/9=4/45$$.