Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 38

Determine the $\displaystyle \lim_{x \to a^+} \frac{\cos x \ln (x - a)}{\ln \left( e^x - e^a\right)}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.


$
\begin{equation}
\begin{aligned}
\lim_{x \to a^+} \frac{\cos x \ln (x - a)}{\ln \left( e^x - e^a\right)} &= \lim_{x \to a^+} \cos x \cdot \lim_{x \to a^+} \left[ \frac{\ln(x-a)}{\ln(e^x - e^a)} \right]\\
\\
&= \cos a \cdot \lim_{x \to a^+} \left[ \frac{\ln(x-a)}{\ln \left( e^x - e^a\right)} \right]
\end{aligned}
\end{equation}
$


By applying L'Hospital's Rule..

$
\begin{equation}
\begin{aligned}
&= \cos a \cdot \lim_{x \to a^+} \left[ \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} \right]\\
\\
&= \cos a \cdot \lim_{x \to a^+} \left( \frac{e^x - e^a}{e^x(x-a)} \right)\\
\\
&= \cos a \left[ \lim_{x \to a^+}\left( \frac{1}{e^x} \right) \cdot \lim_{x \to a^+} \left( \frac{e^x - e^a}{x -a} \right)\right]\\
\\
&= \frac{\cos a}{e^a} \cdot \lim_{x \to a^+} \left( \frac{e^x - e^a}{x - a} \right)
\end{aligned}
\end{equation}
$


Again, by applying L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
&= \frac{\cos a}{e^a} \cdot \lim_{x \to a^+} \left( \frac{e^x}{1} \right)\\
\\
&= \frac{\cos a}{e^a} \cdot \lim_{x \to a^+} e^x\\
\\
&= \frac{\cos a}{e^a} \cdot e^a \quad = \cos a
\end{aligned}
\end{equation}
$