Evaluate $\lim\limits_{x \to a} \; \dfrac{\log \left(x-a\right)}{\log \left(e^x - e^a\right)}$
$\begin{aligned} & \lim\limits_{x \to a} \; \dfrac{\log \left(x-a\right)}{\log \left(e^x - e^a\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx} \left[\log \left(x-a\right)\right]}{\dfrac{d}{dx} \left[\log \left(e^x - e^a\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{1}{x-a}}{\dfrac{e^x}{e^x - e^a}} \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{e^x - e^a}{e^x \left(x-a\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx}\left(e^x - e^a\right)}{\dfrac{d}{dx}\left[e^x \left(x-a\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{e^x}{e^x \left(x-a\right)+e^x} \\ & & \\ & = \dfrac{e^a}{e^a} = 1 \end{aligned}$