Problem 560 of Demidovich's book of problems. Need a solution for a problem 560 from Demidovich's book of problems.
$$
\lim_{x\rightarrow a}\frac{a^{a^x}-a^{x^a}}{a^x-x^a}
$$
I know an answer:
$$
a^{a^a}\ln(a);
$$
I had tired some replacements and some decomposition, but it's not worked.
P.S.
Level of problem mean a solution without l'Hospital.
 A: Using l'Hospitals rule you get
$$
\lim_{x\to a}\frac{a^{a^x}a^x(\ln a)^2-a^{x^a}ax^{a-1}\ln a}{a^x\ln a - ax^{a-1}}=a^{a^a}\ln a.
$$
EDIT: If you're looking for method that does not involve taking derivatives, I suggest the following:
$$\lim_{x\to a}\frac{a^{a^x}-a^{x^a}}{a^x-x^a}=\lim_{x\to a}a^{a^x}\frac{1-a^{x^a-a^x}}{a^x-x^a}=a^{a^a}\underbrace{\lim_{h\to 0}\frac{a^h-1}{h}}_{=\ln a}.$$
In this way you only need $\displaystyle \lim_{h\to 0}\frac{a^h-1}{h}=\ln a$, $\displaystyle\lim_{x\to a}a^x-x^a=0$ and to set $h:=a^x-x^a$.
A: Let $f(t) = a^t=e^{(\ln a) t}$. Now $f'(t) = (\ln a) (a^t)$. 
By the mean value theorem, 
$$\frac{f(t)-f(s) }{ t-s} = f'(c)$$
for some intermediate $c$.
In our case $t$ and $s$ are functions of $x$, satisfying $\lim_{x\to a} t(x) =\lim_{x\to a} s(x) = a^a$. Therefore, the intermediate point $c=c(x)$  tends to $a^a$. Since the derivative of $f$ is continuous, the answer is $f'(a^a)= (\ln a) a^{a^a} $. 
A: The expression $$\frac{a^{a^x} - a^{x^a}} {a^x-x^a} $$ can be written as $$\frac{\exp(a^x\log a) - \exp(x^{a} \log a) }{a^x-x^a}=\exp(x^a\log a) \cdot\frac{\exp((a^x-x^a) \log a) - 1} {(a^x-x^a)\log a} \cdot\log a$$ and the above clearly tends to $\exp(a^a\log a) \log a=a^{a^a} \log a$. 
