The correct answer is $\boxed{1}$.
We can use L’Hôpital’s rule to evaluate the limit. 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
$$\begin{align}
\lim_{x\to 0} \left( {\frac{{{{\text{e}}^{2{\text{x}}}} – 1}}{{\sin \left( {4{\text{x}}} \right)}}} \right) &= \lim_{x\to 0} \left( {\frac{{2{{\text{e}}^{2{\text{x}}}}}}{{4{\text{x}} \cos \left( {4{\text{x}}} \right)}}} \right) \
&= \lim_{x\to 0} \left( {\frac{{{{\text{e}}^{2{\text{x}}}}}}{{4{\text{x}}}}} \right) \cdot \lim_{x\to 0} \left( {\frac{{1}}{{\cos \left( {4{\text{x}}} \right)}}} \right) \
&= 1 \cdot \lim_{x\to 0} \left( {\frac{{1 – \frac{{{x}^{2}}}{2!} + \frac{{{x}^{4}}}{4!} – \cdots }}{{4{\text{x}} – \frac{8{{x}^{3}}}{3!} + \frac{32{{x}^{5}}}{5!} – \cdots }}} \right) \
&= 1 \cdot \lim_{x\to 0} \left( {\frac{{1 – \frac{{{x}^{2}}}{2!} + \frac{{{x}^{4}}}{4!} – \cdots }}{{4 – \frac{8{{x}^{2}}}{3!} + \frac{32{{x}^{4}}}{5!} – \cdots }}} \right) \cdot \lim_{x\to 0} \left( {\frac{{1}}{{4 – \frac{8{{x}^{2}}}{3!} + \frac{32{{x}^{4}}}{5!} – \cdots }}} \right) \
&= 1 \cdot 1 \cdot \lim_{x\to 0} \left( {\frac{{1 – \frac{{{x}^{2}}}{2!} + \frac{{{x}^{4}}}{4!} – \cdots }}{{1 – \frac{2{{x}^{2}}}{3!} + \frac{8{{x}^{4}}}{5!} – \cdots }}} \right) \
&= 1 \cdot 1 \cdot \frac{{1 – \frac{0}{2!}}}{{1 – \frac{0}{3!}}} \
&= 1 \cdot 1 \cdot \frac{{1}}{{1}} \
&= 1.
\end{align}$$
Therefore, the limit $\lim_{x\to 0} \left( {\frac{{{{\text{e}}^{2{\text{x}}}} – 1}}{{\sin \left( {4{\text{x}}} \right)}}} \right)$ is equal to $1$.