Calculate $\lim\limits_{x\to a}\frac{a^{a^{x}}-{a^{x^{a}}}}{a^x-x^a}$

Since this is calculus why not try with L'Hospital?

$$\lim_{x\to a}\frac{a^{a^x}-a^{x^a}}{a^x-x^a}=\lim_{x\to a}\frac{a^xa^{a^x}\log^2a-ax^{a-1}\log a\cdot a^{x^a}}{a^x\log a-ax^{a-1}}=$$

$$=\frac{a^aa^{a^a}\log^2a-a^a\log a\cdot a^{a^a}}{a^a\log a-a^a}=\frac{a^{a^a}\log^2a-\log a\cdot a^{a^a}}{\log a -1}=a^{a^a}\log a$$

Let $f(u)=a^u$. Then the expression inside the limit is $\frac{f(z)-f(y)}{z-y}$ where $z=a^x$ and $y=x^a$.

So, by the mean value theorem, this expression is $f'(c)$ for some $c$ between $z$ and $y$.

But as $x\to a$, both $z$ and $y$ approach $a^a$, and therefore $c$ approaches $a^a$. Since $f'(u)$ is continuous, this means that the limit must be $f'(a^a)$.

So you just need to know $f'(u) = a^u\log a$.

$\displaystyle\lim_{x\to a}\frac{a^{a^{x}}-{a^{x^{a}}}}{a^x-x^a}$=$\displaystyle\lim_{x\to a}\frac{a^{x^{a}}({a^{a^{x}-x^a}}-1)}{a^x-x^a}$=$a^{a^{a}}\displaystyle\lim_{x\to a}\frac{{a^{a^{x}-x^a}-1}}{a^x-x^a}$=$|\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\ln a|$=$a^{x^{a}}\ln a$.