Question:
Evaluate the following limits:
$\lim _{h \rightarrow 0} \frac{\left(e^{3+x}-\sin x-e^{3}\right)}{x}$
Solution:
$=\lim _{x \rightarrow 0} \frac{\left(e^{3+x}-\sin x-e^{3}\right)}{x}$
$=-\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{e^{3+x}-e^{3}}{x}$
$=-1+\lim _{x \rightarrow 0} \frac{e^{3}\left(e^{x}-1\right)}{x}$
$=-1+e^{3}$
$\therefore \lim _{x \rightarrow 0} \frac{\left(e^{3+x}-\sin x-e^{3}\right)}{x}=e^{3}-1$
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