\displaystyle
\lim_{x\rightarrow 0}\frac {x-\sin x} {x^3} = 
\lim_{x\rightarrow 0} ({x\over {x^3}} - {\sin x\over  {x^3}}) =
\lim_{x\rightarrow 0} ({1\over {x^2}} - {{1\over  {x^2}}) = 0