l-hopital-rule

$$\mathop {Lt}\limits_{x \to 0} {{{{\log }_e}\left( {1 + 4x} \right)} \over {{e^{3x}} - 1}}$$ is equal to

$$\mathop {Lim}\limits_{x \to 0} {{{e^x} - \left( {1 + x + {{{x^2}} \over 2}} \right)} \over {{x^3}}} = $$