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

\(\mathop {{\rm{Lt}}}\limits_{x \to 0} \frac{{{{\log }_e}\left( {1 + 4

1.	\(\mathop {{\rm{Lt}}}\limits_{x \to 0} \frac{{{{\log }_e}\left( {1 + 4x} \right)}}{{{e^{3x}} - 1}}\) is equal to
A.	0
B.	\(\frac{1}{{12}}\)
C.	\(\frac{4}{3}\)
D.	1
Answer» D. 1