For what value(s) of $p$, if any, do we have $$ \lim_{x \to p} \frac{ p^x - x^p }{ x^x - p^p } = 1? $$
My Attempt:
Since $0^0$ is indeterminate, we must have $p \neq 0$.
If we put $x = p$ into $\frac{ p^x - x^p }{ x^x - p^p }$, we find that $$ \frac{ p^x - x^p }{ x^x - p^p } = \frac{ p^p - p^p }{ p^p - p^p } = \frac{0}{0}. $$ Thus we have the $\frac{0}{0}$ form. So, provided that the indicated limits exist, we have
$$ \begin{align} \lim_{x \to p} \frac{ p^x - x^p }{ x^x - p^p } &= \lim_{x \to p} \frac{ \frac{d}{dx} \left( p^x - x^p \right) }{ \frac{d}{dx} \left( x^x - p^p \right) } \\ &= \lim_{x \to p} \frac{ p^x \ln p - p x^{p-1} }{ x^x (1 + \ln x) } \tag{1} \\ &= \frac{ p^p ( \ln p - 1 ) }{ p^p ( \ln p + 1) } \\ & \qquad \qquad \mbox{ [ provided that $p \neq e^{-1}$] } \\ &= \frac{ \ln p - 1}{ \ln p + 1}. \end{align} $$ However, if $$ \frac{ \ln p - 1}{ \ln p + 1} = 1, $$ then we also have $$ \ln p - 1 = \ln p + 1, $$ and hence $-1 = 1$, which is impossible.
So let us suppose that $p = e^{-1}$. Then using (1) above, we have $$ \begin{align} \lim_{x \to e^{-1} } \frac{ (1/e)^x - x^{1/e} }{ x^x - e^{-1/e} } &= \lim_{x \to e^{-1} } \frac{ e^{-x} (-1) - e^{-1} x^{ 1/e - 1} }{ x^x (1 + \ln x) } \\ &= \frac{ -e^{-1/e} - e^{-1} \left( e^{-1} \right)^{ 1/e - 1} }{ (1/e)^{1/e} \left(1 + \ln e^{-1} \right) } \\ &= \frac{ -2 e^{-1/e} }{ 0 } \\ &= -\infty. \end{align} $$
Hence there is no value of $p$ for which the indicated limit equals $1$.
Is this solution correct? If not, then where have I erred?
How to solve this problem with full detail and rigor typical of mathematical analysis arguments?