Evaluate the following limits:

$=\lim _{x \rightarrow 0} \frac{\left(e^{3+x}-\sin x-e^{3}\right)}{x}$

$=-\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{e^{3+x}-e^{3}}{x}$

$=-1+\lim _{x \rightarrow 0} \frac{e^{3}\left(e^{x}-1\right)}{x}$

$=-1+e^{3}$

$\therefore \lim _{x \rightarrow 0} \frac{\left(e^{3+x}-\sin x-e^{3}\right)}{x}=e^{3}-1$