Explanation

(a)
$\lim_{x \rightarrow 0}\frac{e^{5x} - e^{4x}}{x}$
$= \lim_{x \rightarrow 0}\frac{\left\lbrack \begin{array}{r}
\left( 1 + \frac{5x}{1!} + \frac{(5x)^{2}}{2!} + \ldots \right) - \\
\left( 1 + \frac{4x}{1!} + \frac{(4x)^{2}}{2!} + ... \right)
\end{array} \right\rbrack}{x}$
$= \lim_{x \rightarrow 0}\frac{x\left\lbrack \begin{array}{r}
\left( \frac{5}{1!} + \frac{25x}{2!} + \ldots \right) - \\
\left( \frac{4}{1!} + \frac{16}{2!} + ... \right)
\end{array} \right\rbrack}{x}$
$= 1$