What is the answer to this math question?

Use Taylor's power series for ${\tan}^{-} 1 x$ to evaluate: ${\lim}_{x \to 0} \frac{{\tan}^{-} 1 x - x}{x} ^ 3$

Jun 17, 2018

${\lim}_{x \to 0} \frac{\arctan x - x}{x} ^ 3 = - \frac{1}{3}$

Explanation:

We want to find ${\lim}_{x \to 0} \frac{{\tan}^{-} 1 x - x}{x} ^ 3 = 0$ using Taylor series

First, let ${\tan}^{-} 1 \equiv \arctan$

Now,

$\arctan x = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7$

so

${\lim}_{x \to 0} \frac{\arctan x - x}{x} ^ 3$

$= {\lim}_{x \to 0} \frac{x - {x}^{3} / 3 + {x}^{5} / 5 - \ldots - x}{x} ^ 3$

$= {\lim}_{x \to 0} \left(- \frac{1}{3} + {x}^{2} / 5 - {x}^{4} / 7. . .\right)$

$= - \frac{1}{3}$