Answer
$cos(a)$
Work Step by Step
$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{ln(x-a)}{ln(e^x - e^a)} =$
We get $\infty/\infty$, so we apply L'Hospital's Rule:
$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} =$
$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{1}{x-a} \times \frac{e^x - e^a}{e^x} =$
$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{e^x - e^a}{x-a} \times \frac{1}{e^x} =$
$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{e^x -
e^a}{x-a} \times \lim\limits_{x \to a^{+}} \frac{1}{e^x} =$
$cos(a) \times \lim\limits_{x \to a^{+}} \frac{e^x - e^a}{x-a} \times \frac{1}{e^a} =$
We get $0/0$, so we apply L'Hospital's Rule:
$cos(a) \times \lim\limits_{x \to a^{+}} \frac{e^x}{1} \times \frac{1}{e^a} =$
$cos(a) \times \lim\limits_{x \to a^{+}} e^a \times \frac{1}{e^a} = cos(a)$
