Explanation

(d)
$\lim_{x \rightarrow 1}\frac{\sin{(e^{x - 1} - 1)}}{\log x}$
$= \lim_{h \rightarrow 0}\frac{\sin{(e^{h} - 1)}}{\log{(1 + h)}}$
$= \lim_{h \rightarrow 0}{\frac{\sin{(e^{h} - 1)}}{(e^{h} - 1)} \times \frac{(e^{h} - 1)}{\log{(1 + h)}}}$
$= 1 \times \lim_{h \rightarrow 0}\frac{\left( h + \frac{h^{2}}{2!} + ... \right)}{\left( h - \frac{h^{2}}{2!} + ...\infty \right)} = 1 \times 1 = 1$