If \(\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}\) = -1 n, what is value of a?

Let, y = x^x
Thus, log y = x log x
Differentiating both sides with respect to x, we get,
1/y dy/dx = (x*1/x) + log x
=>dy/dx = y(1 + log x)
Or, dx^x/dx = x^x(1 + log x)
Using, L?Hospital?s rule,
\(\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}\) = \(\lim\limits_{x \rightarrow a}\frac{(a^x*log?a-x^{a-1})}{x^x(1+logx)}\)
= \(\lim\limits_{x \rightarrow a}\frac{(a^a*log?a-a^{a})}{a^a(1+loga)}\)
= (log a ? 1)/(log a + 1)
As per the question,
(log a ? 1)/(log a + 1) = -1
Or, (log a ? 1) = -log a ? 1
Or, 2 log a = 0
Or, log a = 0
So, a = 1