## Calculus: Early Transcendentals 8th Edition

$cos(a)$
$\lim\limits_{x \to a^{+}} cos(x) \times \lim\limits_{x \to a^{+}} \frac{ln(x-a)}{ln(e^x - e^a)} =$ 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: $cos(a) \times \lim\limits_{x \to a^{+}} \frac{e^x}{1} \times \frac{1}{e^a} =$ $cos(a) \times \lim\limits_{x \to a^{+}} e^a \times \frac{1}{e^a} = cos(a)$