\[\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{{{\log }_{\text{e}}}\left( {1 + 4{\text{x}}} \right)}}{{{{\text{e}}^{3{\text{x}}}} – 1}}\] is equal to A. 0 B. \[\frac{1}{{12}}\] C. \[\frac{4}{3}\] D. 1

The correct answer is $\frac{1}{12}$.

To solve this limit, we can use L’Hôpital’s rule. L’Hôpital’s rule states that if $\lim_{x\to a}\frac{f(x)}{g(x)}$ exists and $\lim_{x\to a}\frac{f'(x)}{g'(x)}$ also exists, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$.

In this case, we have

$$\begin{align}
\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{{{\log }{\text{e}}}\left( {1 + 4{\text{x}}} \right)}}{{{{\text{e}}^{3{\text{x}}}} – 1}} &= \mathop {\lim }\limits{{\text{x}} \to 0} \frac{{\frac{d}{d{\text{x}}}{{\log }{\text{e}}}\left( {1 + 4{\text{x}}} \right)}}{{\frac{d}{d{\text{x}}}\left( {{\text{e}}^{3{\text{x}}}} – 1 \right)}} \
&= \mathop {\lim }\limits{{\text{x}} \to 0} \frac{{\frac{4}{{1 + 4{\text{x}}}}}}{{3{{\text{e}}^{3{\text{x}}}} \cdot 3{\text{x}}}} \
&= \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{4}}{{3{{\text{e}}^{3{\text{x}}}} \cdot {{\left( 1 + 4{\text{x}} \right)}^{2}}} \
&= \frac{4}{{3 \cdot 1 \cdot 1}} \
&= \frac{1}{{12}}.
\end{align}$$

Therefore, the limit $\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{{{\log }_{\text{e}}}\left( {1 + 4{\text{x}}} \right)}}{{{{\text{e}}^{3{\text{x}}}} – 1}}$ is equal to $\frac{1}{12}$.

The other options are incorrect because they do not equal $\frac{1}{12}$.