Answer
$\dfrac{-1}{3}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{\tan^{-1} x-x}{x^3}$
Recall the series for $\tan^{-1} x=\Sigma_{k =0}^{\infty} (-1)^k \dfrac{x^{2k+1}}{2k+1}$
Now, $L=\lim\limits_{x \to 0} \dfrac{\tan^{-1} x-x}{x^3} \\=\lim\limits_{x \to 0} \dfrac{\Sigma_{k =0}^{\infty} (-1)^k \dfrac{x^{2k+1}}{2k+1}-x}{x^3} \\=\lim\limits_{x \to 0}(\Sigma_{k =0}^{\infty}(-1)^k \dfrac{x^{2k-2}}{2k+1}-\dfrac{x}{x^3}) \\=\lim\limits_{x \to 0} (\dfrac{1}{x^2}+\Sigma_{k =1}^{\infty} (-1)^k \dfrac{x^{2k-2}}{2k+1}-\dfrac{1}{x^2})\\=\lim\limits_{x \to 0} [\Sigma_{k =1}^{\infty} (-1)^k \dfrac{x^{2k-2}}{2k+1}]\\=\lim\limits_{x \to 0} [(-1)^1 \dfrac{x^{2(1)-2}}{2(1)+1}+(-1)^2 \dfrac{x^{2(2)-2}}{2(2)+1}+......]\\=-\dfrac{1}{3}+\dfrac{0^2}{5}-\dfrac{0^4}{7}+........\\=\dfrac{-1}{3}$
