Answer
$\infty$
Work Step by Step
$\lim _{x^+\rightarrow 0}\dfrac {e^{x}-\left( 1+x\right) }{x^{3}}=\dfrac {\dfrac {d}{dx}\left( ex-\left( 1+x\right) \right) }{\dfrac {d}{dx}\left( x^{3}\right) }=\dfrac {e^{x}-1}{3x^{2}}=\dfrac {\dfrac {d}{dx}\left( e^x-1\right) }{\dfrac {d}{dx}\left( 3x^{2}\right) }=\dfrac {e^{x}}{6x}=\dfrac {e^{0}}{0}=\infty $
