If $\lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} = -1$, find the value of a.
$\begin{aligned}
& \lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} & \left(\dfrac{0}{0} \text{ form}\right) \\
& & \\
& = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx}\left(a^x - x^a\right)}{\dfrac{d}{dx}\left(x^x - a^a\right)} & \left[\text{L'Hospital's rule}\right] \\
& & \\
& = \lim\limits_{x \to a} \; \dfrac{a^x \ln a - ax^{a-1}}{x^x \left(1+ \ln x\right)} & \\
& & \\
& = \dfrac{a^a \ln a -a \times a^{a-1}}{a^a \left(1+ \ln a\right)} & \\
& & \\
& = \dfrac{a^a \ln a -a^a}{a^a \left(1+ \ln a\right)} & \\
& & \\
& = \dfrac{\ln a -1}{\ln a + 1} &
\end{aligned}$
$\therefore$ $\;$ $\lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} = -1$ $\implies$ $\dfrac{\ln a - 1}{\ln a + 1}=-1$
i.e. $\ln a -1 = -\ln a - 1$
i.e. $2 \ln a = 0$
i.e. $a = e^0 = 1$