[tex]\lim\limits_{x\to 0}\dfrac{x-\arctan x}{x^3} \overset{\frac{0}{0}(L'H)}{=} \lim\limits_{x\to 0}\dfrac{1-\dfrac{1}{1+x^2}}{3x^2}=\lim\limits_{x\to 0}\dfrac{1+x^2-1}{3x^2(1+x^2)}= \\ \\ =\lim\limits_{x\to 0}\dfrac{1}{3(1+x^2)}= \dfrac{1}{3(1+0)} = \boxed{\dfrac{1}{3}}[/tex]
